Factor Formulae - Derivation

$ \displaystyle \begin{array}{l}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta ---(1)\\\\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta ---(2)\ \ \ \end{array}$

ဆိုတဲ့ Sum and Difference Formulae ေတြ ကို မွတ္မိၾကမယ္ ထင္ပါတယ္။

ညီမွ်ျခင္း (1) နဲ႔ (2) ကို ေပါင္းလိုက္မယ္...။

$ \displaystyle \sin (\alpha +\beta )+\ \sin (\alpha -\beta )=2\sin \alpha \cos \beta ---(*)$

ဆိုၿပီး ရလာမွာေပါ့...။

$ \displaystyle \alpha +\beta =\theta $ နဲ႔ $ \displaystyle \alpha -\beta =\phi $ လို႔ ထားလိုက္မယ္။

ဒါဆိုရင္ $ \displaystyle \alpha =\frac{{\theta +\phi }}{2}$ နဲ႔ $ \displaystyle \beta =\frac{{\theta -\phi }}{2}$ ျဖစ္သြားမွာေပါ့...။

$ \displaystyle \alpha +\beta =\theta ,\ \alpha -\beta =\phi ,\ \ \alpha =\frac{{\theta +\phi }}{2},\beta =\frac{{\theta -\phi }}{2}$ တို႔ကို (*) မွာ အစားသြင္းလိုက္တဲ့ အခါ မွာေတာ့ ေအာက္ပါ factor formula ကို ရရွိမွာ ျဖစ္ပါတယ္။

$ \displaystyle \sin \theta +\ \sin \phi =2\sin \frac{{\theta +\phi }}{2}\cos \frac{{\theta -\phi }}{2}$

ဒီတစ္ခါ ညီမွ်ျခင္း (1) ထဲက (2) ကို ႏႈတ္ပါမယ္...။

$ \displaystyle \sin (\alpha +\beta )-\ \sin (\alpha -\beta )=2\cos \alpha \sin \beta $

အထက္ကအတိုင္း သက္ဆိုင္ရာတန္ဖိုးေတြ အစားသြင္းလိုက္ရင္ ...။

$ \displaystyle \sin \theta -\ \sin \phi =2\cos \frac{{\theta +\phi }}{2}\sin \frac{{\theta -\phi }}{2}$

Identity တစ္ခု ထပ္ရပါမယ္...။

ေနာက္ထပ္ညီမွ်ျခင္း ႏွစ္ၾကာင္း ကို ဆက္ၾကည့္ရေအာင္...။

$ \displaystyle \begin{array}{l} \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta ---(3)\\\\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta ---(4)\ \ \ \end{array}$

အထက္ပါအတိုင္း ညီမ ွ်ျခင္း ႏွစ္ေၾကာင္း (3) နဲ႔ (4) ကို ေပါင္းတစ္လွည့္ ႏႈတ္တစ္လွည့္ လုပ္လိုက္ရင္ ...။

$ \displaystyle \begin{array}{l}(3)+(4)\Rightarrow \ \ \ \cos (\alpha +\beta )+\ \cos (\alpha -\beta )=2\cos \alpha \cos \beta \\\\(3)-(4)\Rightarrow \ \ \ \cos (\alpha +\beta )-\ \cos (\alpha -\beta )=-2\cos \alpha \cos \beta \end{array}$

အထက္မွာ ရွာခဲ့ၿပီး ျဖစ္တဲ့ $ \displaystyle \alpha +\beta =\theta ,\ \alpha -\beta =\phi ,\ \ \alpha =\frac{{\theta +\phi }}{2},\beta =\frac{{\theta -\phi }}{2}$ တို႔ကို သက္ဆိုင္ရာ တန္ဖိုးေတြမွာ အစားသြင္းလိုက္ရင္...။

$ \displaystyle \begin{array}{l}\cos \theta +\ \cos \phi =2\cos \displaystyle \frac{{\theta +\phi }}{2}\cos \displaystyle \frac{{\theta -\phi }}{2}\\\\\cos \theta -\ \cos \phi =-2\sin \displaystyle \frac{{\theta +\phi }}{2}\sin \displaystyle \frac{{\theta -\phi }}{2}\end{array}$

Half-Angle Formulae - Derivation

$ \displaystyle \ \cos 2\alpha =1-2{{\sin }^{2}}\alpha $ ဆိုတာ သိခဲ့ပါၿပီ။

$ \displaystyle 2\alpha = \theta$ လို႔ ထားလိုက္မယ္။ ဒါဆိုရင္ $ \displaystyle \theta =\frac{\alpha}{2}$ ေပါ့...။

မူလညီမွ်ျခင္းမွာ အစားသြင္းလိုက္ ရင္ ...

$ \displaystyle \begin{array}{l}2{{\sin }^{2}}\displaystyle \frac{\theta }{2}=1-\cos \theta \\\\{{\sin }^{2}}\displaystyle \frac{\theta }{2}=\displaystyle \frac{{1-\cos \theta }}{2}\end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \sin \frac{\theta }{2}=\pm \sqrt{{\frac{{1-\cos \theta }}{2}}}$

$ \displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$ လို႔လည္း သိထားခဲ့ၿပီးသား မဟုတ္လား . . .။

အထက္မွာ ေျပာခဲ့တဲ့အတိုင္း $ \displaystyle 2\alpha =\theta ,\alpha =\frac{\theta }{2}$ ကို အစားသြင္းလိုက္ရင္ ...

$ \displaystyle \begin{array}{l}2{{\cos }^{2}}\displaystyle \frac{\theta }{2}=1+\cos \theta \\\\{{\cos }^{2}}\displaystyle \frac{\theta }{2}=\displaystyle \frac{{1+\cos \theta }}{2}\end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \cos \frac{\theta }{2}=\pm \sqrt{{\frac{{1+\cos \theta }}{2}}}$

$ \displaystyle \sin \frac{\theta }{2}$ နဲ႕ $ \displaystyle \cos \frac{\theta }{2}$ ကို သိၿပီဆိုေတာ့ ....

$ \displaystyle \tan \displaystyle \frac{\theta }{2}= \displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}$ ဆိုတဲ့ basic identity ကို သံုးၿပီး ဆက္ရွာလို႔ရၿပီေပါ့။

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \displaystyle \frac{{\sqrt{{\displaystyle \frac{{1-\cos \theta }}{2}}}}}{{\sqrt{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{\displaystyle \frac{{1-\cos \theta }}{2}}}{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}}}$

ဆက္ၿပီး derive လုပ္ၾကည့္မယ္ ...။

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1+\cos \theta }}{{1+\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-{{{\cos }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \frac{\theta }{2}=\frac{{\sin \theta }}{{1+\cos \theta }}$

တဖန္ . . .

$ \displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1-\cos \theta }}{{1-\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{1-{{{\cos }}^{2}}\theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{{{{\sin }}^{2}}\theta }}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

$ \displaystyle \tan \frac{\theta }{2}=\frac{{1-\cos \theta }}{{\sin \theta }}$

Double Angle Formulae - Derivation

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ ဆိုတာ သိခဲ့ၿပီး ျဖစ္မယ္ ထင္ပါတယ္။

ဒီ ပံုေသနည္းဟာ မည္သည့္ေထာင့္ $ \displaystyle \alpha$ နဲ႔ $ \displaystyle \beta$ အတြက္မဆို မွန္ပါတယ္။

ဒါဆိုရင္ $ \displaystyle \alpha=\beta$ အတြက္လည္း မွန္တာေပါ့။... ဒါေၾကာင့္

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \sin (\alpha +\alpha )=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha $

ဒါ့ေၾကာင့္

$ \displaystyle \sin 2\alpha =2\sin \alpha \cos \alpha $

အလားတူပါပဲ.....။

$ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \cos (\alpha +\alpha )=\cos \alpha \cos \alpha -\sin \beta \sin \beta $

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $

$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ဆိုတဲ့ Pythagorean Identity ကို မွတ္မိမယ္ ထင္ပါတယ္။

$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ျဖစ္တာေၾကာင့္ $ \displaystyle {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha $ နဲ႔ $ \displaystyle {{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha $ ျဖစ္ပါတယ္။

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $ ဆိုတဲ့ equation မွာ သက္ဆိုင္ရာ တန္ဖိုးေတြကို အစားသြင္းလိုက္ရင္ ...

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha =1-{{\sin }^{2}}\alpha -{{\sin }^{2}}\alpha \end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =1-2{{\sin }^{2}}\alpha $

အလားတူပါပဲ...။

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha ={{\cos }^{2}}\alpha -(1-{{\cos }^{2}}\alpha )\end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$

$ \displaystyle \tan 2\alpha $ အတြက္ ဆက္ရွာၾကည့္ပါမယ္။

$ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$ လို႔ သိခဲ့ၿပီးပါၿပီ။..

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \tan (\alpha +\alpha )=\frac{{\tan \alpha +\tan \alpha }}{{1-\tan \alpha \tan \alpha }}$ ..

ဒါ့ေၾကာင့္...

$ \displaystyle \tan 2\alpha =\frac{{2\tan \alpha }}{{1-{{{\tan }}^{2}}\alpha }}$

Sum and Difference Formulae - Derivation

$ \displaystyle ΔABC, ΔACD$ နဲ႔ $ \displaystyle ΔCDF$ တို႔ဟာ ေထာင့္မွန္ႀတိဂံမ်ား ျဖစ္ၾကပါတယ္။

$ \displaystyle ΔABC$ မွာ $ \displaystyle ∠CAB$ ကို $ \displaystyle \alpha$ လို႔ သတ္မွတ္ပါမယ္။

$ \displaystyle ΔABC\simΔCDF$ ျဖစ္တာေၾကာင့္ $ \displaystyle ∠CDF=\alpha$ ျဖစ္ပါတယ္။

$ \displaystyle ΔACD$ မွာေတာ့ $ \displaystyle ∠CAD$ ကို $ \displaystyle \beta$ လို႔ သတ္မွတ္ပါမယ္။

ဒါဆိုရင္ $ \displaystyle ΔABC$ မွာ...

$ \displaystyle \sin \alpha=\frac{BC}{AC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{AB}{AC}$ ျဖစ္ပါတယ္။

ဒါ့ေၾကာင့္ $ \displaystyle BC =AC \sin \alpha$ နဲ႕ $ \displaystyle AB =AC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

တဖန္ $ \displaystyle ΔCDF$ မွာ...

$ \displaystyle \sin \alpha=\frac{FC}{DC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{DF}{DC}$ ျဖစ္ပါတယ္။

ဒီမွာလည္း $ \displaystyle FC =DC \sin \alpha$ နဲ႕ $ \displaystyle DF =DC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

$ \displaystyle ΔACD$ မွာလည္း ...

$ \displaystyle \sin \beta=\frac{DC}{AD}$ နဲ႕ $ \displaystyle \cos \beta=\frac{AC}{AD}$ ျဖစ္ပါတယ္။

ဒါဆိုရင္ $ \displaystyle DC =AD \sin \alpha$ နဲ႕ $ \displaystyle AC =AD \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။

$ \displaystyle ΔADE$ အတြက္ ဆက္ၾကည့္ရေအာင္...

$ \displaystyle \sin (\alpha+\beta)=\frac{DE}{AD}$ ျဖစ္ပါတယ္။

ပံုမွာ ေတြ႔ရတဲ့ အတိုင္း $ \displaystyle DE =DF+FE$ ျဖစ္ပါတယ္။။

ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{FE}}{{AD}}$ လို႔ ေျပာႏိုင္ပါတယ္။ ။

တဖန္ ့ $ \displaystyle BCFE$ က rectangle ျဖစ္တာေၾကာင့္ $ \displaystyle FE=BC$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။ ။

ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{BC}}{{AD}}$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။။

$ \displaystyle DF =DC \cos \alpha, BC =AC \sin \alpha$ လို႔ အထက္မွာ သိခဲ့ၿပီးပါၿပီ။ ဒါဆိုရင္ ။

$ \displaystyle \sin (\alpha +\beta )=\frac{{DC}}{{AD}}\cos \alpha +\frac{{AC}}{{AD}}\sin \alpha $ လို႔ ေျပာလို႔ရတာေပါ့။ ။

ဒီအခါမွာလည္း $ \displaystyle \sin \beta=\frac{DC}{AD}, \cos \beta=\frac{AC}{AD}$ လို႕သိခဲ့ၿပီးပါၿပီ။ ဒါေၾကာင့္။

$ \displaystyle \sin (\alpha +\beta )=\sin \beta \cos \alpha +\cos \beta \sin \alpha $ လို႔ ေျပာလို႔ရပါၿပီ။ ျပန္စီလိုက္ရင္ ...။

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $

အထက္ပါအတိုင္း ... $ \displaystyle \cos (\alpha +\beta )$ အတြက္ ပံုေသနည္းကို ဆက္ရွာႏိုင္ပါတယ္။ ..။

$ \displaystyle \begin{array}{l}\cos (\alpha +\beta )=\displaystyle \frac{{AE}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-EB}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-FC}}{{AD}}\ \ \ \ \left[ {\because EB=FC} \right]\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB}}{{AD}}-\displaystyle \frac{{FC}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AC}}{{AD}}\cos \alpha -\displaystyle \frac{{DC}}{{AD}}\sin \alpha \\\\\text{Since}\ \displaystyle \frac{{AC}}{{AD}}=\cos \beta \ \operatorname{and}\ \displaystyle \frac{{DC}}{{AD}}=\sin \beta ,\\\\\text{Therefore,}\end{array}$

$ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $

$ \displaystyle \sin (\alpha +\beta )$ နဲ႔ $ \displaystyle \cos (\alpha +\beta )$ ကို သိၿပီဆိုေတာ့ $ \displaystyle \tan (\alpha +\beta )$ ကို ရွာႏိုင္ၿပီေပါ့။

$ \displaystyle \begin{array}{l}\tan (\alpha +\beta )=\displaystyle \frac{{\sin (\alpha +\beta )}}{{\cos (\alpha +\beta )}}\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta -\sin \alpha \sin \beta }}\\\\\text{Dividing the numerator and denominator }\\\text{with}\ \cos \alpha \cos \beta ,\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\displaystyle \frac{{\sin \alpha \cos \beta }}{{\cos \alpha \cos \beta }}+\displaystyle \frac{{\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\displaystyle \frac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }}-\displaystyle \frac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}\\\\\text{Therefore,}\end{array}$

$ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$

$ \displaystyle \begin{array}{l}\ \ \ \ \sin (-\alpha )=-\sin \alpha ,\\\\\ \ \ \ \cos (-\alpha )=\cos \alpha ,\\\\\ \ \ \ \tan (-\alpha )=-\tan \alpha \\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$

$ \displaystyle \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $

$ \displaystyle \begin{array}{l}\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos (-\beta )-\sin \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$

$ \displaystyle \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $

$ \displaystyle \begin{array}{l}\ \ \ \ \tan \left( {\alpha -\beta } \right)=\tan \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \tan \left( {\alpha -\beta } \right)=\displaystyle \frac{{\tan \alpha +\tan (-\beta )}}{{1-\tan \alpha \tan (-\beta )}}\end{array}$

$ \displaystyle \tan \left( {\alpha -\beta } \right)=\frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}$

အားလံုးျပန္ေပါင္းရရင္....

$ \displaystyle \begin{array}{l} \sin \left( {\alpha \pm \beta } \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\\\ \cos \left( {\alpha \pm \beta } \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\\\ \tan \left( {\alpha \pm \beta } \right)=\displaystyle \frac{{\tan \alpha \pm \tan (-\beta )}}{{1\mp \tan \alpha \tan (-\beta )}}\end{array}$

Exercise (11.3) No(4, 5) - Solution

4. If $ \displaystyle α + β + γ = 180°$, prove that

$ \displaystyle \text{(a)}\ \ \sin (\alpha +\beta )=\cos (90{}^\circ -\gamma )$

$ \displaystyle \text{(b)}\ \ \sin (\frac{{\alpha +\beta }}{2})=\sin (90{}^\circ +\frac{\gamma }{2})$

$ \displaystyle \text{(c)}\ \ \tan \left( {\frac{\alpha }{2}} \right)=\cot \left( {180{}^\circ +\frac{{\beta +\gamma }}{2}} \right)$

Show/Hide Solution

$ \displaystyle \text{(a)}\ \ \alpha +\beta +\gamma =180{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \alpha +\beta =180{}^\circ -\gamma \\\\\therefore \ \ \ \ \sin (\alpha +\beta )=\sin (180{}^\circ -\gamma )\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin \gamma \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cos (90{}^\circ -\gamma )\end{array}$


$ \displaystyle \begin{array}{l}\text{(b)}\ \ \alpha +\beta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \alpha +\beta =180{}^\circ -\gamma \\\\\ \ \ \ \ \ \displaystyle \frac{{\alpha +\beta }}{2}=\displaystyle \frac{{180{}^\circ -\gamma }}{2}=90{}^\circ \displaystyle -\frac{\gamma }{2}\\\\\therefore \ \ \ \ \sin (\displaystyle \frac{{\alpha +\beta }}{2})=\sin (90{}^\circ \displaystyle -\frac{\gamma }{2})\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos \displaystyle \frac{\gamma }{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \sin (90{}^\circ +\displaystyle \frac{\gamma }{2})\end{array}$


$ \displaystyle \begin{array}{l}\text{(c)}\ \ \alpha +\beta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \alpha =180{}^\circ -(\beta +\gamma )\\\\\ \ \ \ \ \ \displaystyle \frac{\alpha }{2}=\displaystyle \frac{{180{}^\circ -(\beta +\gamma )}}{2}=90{}^\circ \displaystyle -\frac{{\beta +\gamma }}{2}\\\\\therefore \ \ \ \ \tan \left( {\displaystyle \frac{\alpha }{2}} \right)=\tan (90{}^\circ \displaystyle -\frac{{\beta +\gamma }}{2})\\\ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cot \displaystyle \frac{{\beta +\gamma }}{2}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cot \left( {180{}^\circ +\displaystyle \frac{{\beta +\gamma }}{2}} \right)\end{array}$

5.    Prove that in any triangle $ \displaystyle ABC,$

(i) $ \displaystyle \sin (A+B) = \sin C.$

(ii) $ \displaystyle \cos(A+B) + \cos C = 0.$

(iii) $ \displaystyle \cos \frac{A+B}{2} = \sin \frac{C}{2}.$

(iv) $ \displaystyle \tan \frac{A+B}{2} = \cot \frac{C}{2}.$

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(i) $ \displaystyle \text{Since}\ A+B+C=180{}^\circ ,$

$ \displaystyle \begin{array}{l}\therefore A+B=180{}^\circ -C\\\\\therefore \sin (A+B)=\sin (180{}^\circ -C)\\\\\therefore \sin (A+B)=\sin C\end{array}$


(ii) $ \displaystyle \text{Similarly, }\cos (A+B)=\cos (180{}^\circ -C)$

$ \displaystyle \begin{array}{l}\therefore \cos (A+B)=-\cos C\\\\\therefore \cos (A+B)+\cos C=0\end{array}$


(iii) $ \displaystyle \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\cos \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \cos \left( {\frac{{A+B}}{2}} \right)=\sin \frac{C}{2}$


(iv) $ \displaystyle \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {\frac{{180{}^\circ -C}}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\tan \left( {90{}^\circ -\frac{C}{2}} \right)$

$ \displaystyle \therefore \ \tan \left( {\frac{{A+B}}{2}} \right)=\cot \frac{C}{2}$

Exercise (11.3) No (2) Solution

Find the value of θ, 0° ≤ θ ≤ 360° for the following equations. Do not use table.

(a) $ \displaystyle \ \ \ \ \ \ \sin \theta =-\frac{1}{2}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\sin \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{3}}^{{\text{rd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\end{array}$

 $ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ +30{}^\circ \ (\text{or)}\ \theta =360{}^\circ -30{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =210{}^\circ \ (\text{or)}\ \theta =330{}^\circ $

(b) $ \displaystyle \ \ \ \ \ \ \cos \theta =-\frac{{\sqrt{3}}}{2}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ -30{}^\circ \ (\text{or)}\ \theta =180{}^\circ +30{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =150{}^\circ \ (\text{or)}\ \theta =210{}^\circ $

(c) $ \displaystyle \ \ \ \ \ \ \cos \theta =-\frac{1}{{\sqrt{2}}}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =180{}^\circ -45{}^\circ \ (\text{or)}\ \theta =180{}^\circ +45{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =135{}^\circ \ (\text{or)}\ \theta =225{}^\circ $

(d) $ \displaystyle \ \ \ \ \ \ \tan \theta =\sqrt{3}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=60{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan \theta \ \text{is positive, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \,\ \ \ \theta =60{}^\circ \ (\text{or)}\ \theta =180{}^\circ +60{}^\circ \ $

$ \displaystyle \therefore \,\ \ \ \theta =60{}^\circ \ (\text{or)}\ \theta =240{}^\circ $

(e) $ \displaystyle \ \ \ \ \ \ \tan 2\theta =1$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan 2\theta \ \text{is positive, }2\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \underline{{\text{For }{{\text{1}}^{{\text{st}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \therefore \,\ \ \ 2\theta =45{}^\circ \ (\text{or)}\ 2\theta =360{}^\circ +45{}^\circ \ $

$ \displaystyle \ \ \ \ \ 2\theta =45{}^\circ \ (\text{or)}\ 2\theta =405{}^\circ \ $

$ \displaystyle \ \ \ \ \ \theta =22{}^\circ 3{0}'(\text{or)}\ \theta =\ 202{}^\circ 3{0}'$

$ \displaystyle \underline{{\text{For }{{\text{3}}^{{\text{rd}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 2\theta =180{}^\circ +45{}^\circ \ (\text{or)}\ 2\theta =360{}^\circ +180{}^\circ +45{}^\circ $

$ \displaystyle \ \ \ \ \ 2\theta =225{}^\circ \ (\text{or)}\ 2\theta =585{}^\circ $

$ \displaystyle \ \ \ \ \ \theta =112{}^\circ 3{0}'(\text{or)}\ \theta =292{}^\circ 3{0}'$

(f) $ \displaystyle \ \ \ \ \ \ \tan 3\theta =-1$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=45{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But }\tan 3\theta \ \text{is negative, 3}\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\end{array}$

$ \displaystyle \underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 3\theta =180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =135{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =45{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}$

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =495{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =165{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ +180{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =855{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =285{}^\circ $

$ \displaystyle \underline{{\text{For }{{\text{4}}^{{\text{th}}}}\ \text{quadrant}}}\text{,}$

$ \displaystyle \ \,\ \ \ 3\theta =360{}^\circ -45{}^\circ $

$ \displaystyle \ \,\ \ \ 3\theta =315{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =105{}^\circ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ -45{}^\circ $

$ \displaystyle \ \,\ \ \ 3\theta =675{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =225{}^\circ \ \ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ $

$ \displaystyle \ \ \ \ \ 3\theta =360{}^\circ +360{}^\circ +360{}^\circ -45{}^\circ \ $

$ \displaystyle \ \,\ \ \ 3\theta =1035{}^\circ \ $

$ \displaystyle \ \,\ \ \ \theta =345{}^\circ \ \ $

diagram မ်ားသည္ principal angle မ်ား ကို ဆံုးျဖတ္ရာတြင္  အေထာက္အကူ ျဖစ္ေစရန္ ေရးဆြဲ ေဖၚျပျခင္း ျဖစ္သည္။ နားလည္ ကၽြမ္းက်င္ သြားလ်င္ diagram မ်ား မေရးဆြဲပဲ ေျဖဆိုႏိုင္ ပါသည္။ က်န္ေသာပုစာၦမ်ားအတြက္ diagram မဆြဲေတာ့ပဲ ေျဖဆိုပါမည္။

(g) $ \displaystyle \ \ \ \ \ \ \tan (3\theta -30{}^\circ )=-1$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=45{}^\circ \\\\\ \ \ \ \ \ \text{But }\tan (3\theta -30{}^\circ )\ \text{is negative, }(3\theta -30{}^\circ )\ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 3\theta -30{}^\circ =180{}^\circ -45{}^\circ \\\ \\\ \,\ \ \ 3\theta =165{}^\circ \ \\\\\ \,\ \ \ \theta =55{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +180{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =525{}^\circ \ \\\\\ \,\ \ \ \theta =175{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ +180{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =885{}^\circ \ \\\\\ \,\ \ \ \theta =295{}^\circ \ \ \ \\\\\underline{{\text{For }{{\text{4}}^{{\text{th}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 3\theta -30{}^\circ =360{}^\circ -45{}^\circ \\\\\ \,\ \ \ 3\theta =345{}^\circ \ \\\\\ \,\ \ \ \theta =115{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ -45{}^\circ \\\\\ \,\ \ \ 3\theta =705{}^\circ \ \\\\\ \,\ \ \ \theta =295{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 3\theta -30{}^\circ =360{}^\circ +360{}^\circ +360{}^\circ -45{}^\circ \ \\\\\ \,\ \ \ 3\theta =1065{}^\circ \ \\\\\ \,\ \ \ \theta =355{}^\circ \ \ \end{array}$

(h) $ \displaystyle \ \ \ \ \ \ \cos 2\theta =-\frac{1}{2}$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=60{}^\circ \\\\\ \ \ \ \ \ \text{But }\cos 2\theta \ \text{is negative, }2\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta =180{}^\circ -60{}^\circ \\\ \\\ \,\ \ \ 2\theta =120{}^\circ \ \\\\\ \,\ \ \ \theta =60{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ -60{}^\circ \ \\\\\ \,\ \ \ 2\theta =480{}^\circ \ \\\\\ \,\ \ \ \theta =240{}^\circ \ \\\\\underline{{\text{For }{{\text{3}}^{{\text{rd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta =180{}^\circ +60{}^\circ \\\\\ \,\ \ \ 2\theta =240{}^\circ \ \\\\\ \,\ \ \ \theta =120{}^\circ \ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ +60{}^\circ \\\\\ \,\ \ \ 2\theta =600{}^\circ \ \\\\\ \,\ \ \ \theta =300{}^\circ \ \ \ \ \end{array}$

(i) $ \displaystyle \ \ \ \ \ \sin (2\theta +30{}^\circ )=\frac{{\sqrt{3}}}{2}$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=60{}^\circ \\\\\ \ \ \ \ \ \text{But }\sin (2\theta +30{}^\circ )\ \text{is positive, }2\theta +30{}^\circ \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\\\\\underline{{\text{For }{{\text{1}}^{{\text{st}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta +30{}^\circ =60{}^\circ \\\ \\\ \,\ \ \ 2\theta =30{}^\circ \ \\\\\ \,\ \ \ \theta =15{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +60{}^\circ \ \\\\\ \,\ \ \ 2\theta =420{}^\circ \ \\\\\ \,\ \ \ \theta =210{}^\circ \ \\\\\underline{{\text{For }{{\text{2}}^{{\text{nd}}}}\ \text{quadrant}}}\text{,}\\\\\ \,\ \ \ 2\theta +30{}^\circ =180{}^\circ -60{}^\circ \\\ \\\ \,\ \ \ 2\theta =120{}^\circ \ \\\\\ \,\ \ \ \theta =60{}^\circ \ \\\\\ \ \ \ \ \ \ \ \ \ \ (\text{or)}\ \\\\\ \ \ \ \ 2\theta =360{}^\circ +180{}^\circ -60{}^\circ \ \\\\\ \,\ \ \ 2\theta =480{}^\circ \ \\\\\ \,\ \ \ \theta =240{}^\circ \ \end{array}$

(j) $ \displaystyle \ \ \ \ \ \ \tan \frac{1}{2}\theta =-\frac{1}{{\sqrt{3}}}$

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$ \displaystyle \therefore \ \ \ \ \text{basic acute angle}=30{}^\circ $

$ \displaystyle \ \ \ \ \ \ \text{But}\ \tan \frac{1}{2}\theta \ \text{is positive, }\frac{1}{2}\theta \ \text{may be }$

$ \displaystyle \ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{4}}^{{\text{th}}}}\text{ quadrant}\text{.}$

$ \displaystyle \therefore \ \ \ \ \frac{1}{2}\theta =180{}^\circ -30{}^\circ \ (\text{or})\ \frac{1}{2}\theta =360{}^\circ -30{}^\circ $

$ \displaystyle \therefore \ \ \ \ \frac{1}{2}\theta =150{}^\circ \ (\text{or})\ \frac{1}{2}\theta =330{}^\circ $

$ \displaystyle \therefore \ \ \ \ \theta =300{}^\circ \ (\text{or})\ \theta =660{}^\circ $

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \text{But}\ 0{}^\circ \le \theta \le 360{}^\circ ,\theta =660{}^\circ \ \text{is not in domain}\text{.}\\\\\therefore \ \ \ \ \theta =300{}^\circ \ \text{is the only solution}\text{.}\end{array}$

(k) $ \displaystyle \ \ \ \ \ \ \sin \theta =0.6521$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=40{}^\circ 4{2}'\ (\text{using table)}\\\\\ \ \ \ \ \ \text{But}\ \sin \theta \ \text{is positive, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{1}}^{{\text{st}}}}\text{ or }{{\text{2}}^{{\text{nd}}}}\text{ quadrant}\text{.}\\\\\therefore \ \ \ \ \theta =40{}^\circ 4{2}'\ (\text{or})\ \theta =180{}^\circ -40{}^\circ 4{2}'\\\\\therefore \ \ \ \ \theta =40{}^\circ 4{2}'\ (\text{or})\ \theta =139{}^\circ 1{8}'\\\ \end{array}$

(l) $ \displaystyle \ \ \ \ \ \ \cos \theta =-0.3854$

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$ \displaystyle \begin{array}{l}\therefore \ \ \ \ \text{basic acute angle}=67{}^\circ 2{0}'\ (\text{using table)}\\\\\ \ \ \ \ \ \text{But}\ \cos \theta \ \text{is negative, }\theta \ \text{may be }\\\ \ \ \ \ \ \text{either in the }{{\text{2}}^{{\text{nd}}}}\text{ or }{{\text{3}}^{{\text{rd}}}}\text{ quadrant}\text{.}\\\\\therefore \ \ \ \ \theta =180{}^\circ -67{}^\circ 2{0}'\ (\text{or})\ \theta =180{}^\circ +67{}^\circ 2{0}'\\\\\therefore \ \ \ \ \theta =112{}^\circ 4{0}'\ (\text{or})\ \theta =247{}^\circ 2{0}'\end{array}$

Exercise (11.3) No (3) Solution

Solve the following equations for 0° ≤ x ≤ 360°.

$ \displaystyle {(\text{a})\ 2\sin x\cos x=\sin x}$

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$ \displaystyle {\ \ \ \ \ 2\sin x\cos x-\sin x=0}$

$ \displaystyle {\ \ \ \ \ \sin x(2\cos x-1)=0}$

$ \displaystyle \ \ \ \ \ \sin x=0\ \ (\text{or})\ \cos x=\frac{1}{2}$

$ \displaystyle \begin{array}{l}(\text{i})\ \ \text{For }\sin x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \end{array}$

$ \displaystyle (\text{ii})\ \text{For }\cos x=\frac{1}{2},$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=360{}^\circ -60{}^\circ \\\\\ \ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=300{}^\circ \end{array}$

$ \displaystyle (\text{b})\ 3\tan x\sin x=2\tan x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3\tan x\sin x-2\tan x=0\\\\\ \ \ \ \ \tan x(3\sin x-2)=0\\\\\ \ \ \ \ \tan x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{2}{3}\\\\(\text{i})\ \ \text{For }\tan x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{2}{3}=0.6667,\\\\\ \ \ \ \ \ x=41{}^\circ 4{9}'\ (\text{or})\ x=180{}^\circ -41{}^\circ 4{9}'{}^\circ \\\\\ \ \ \ \ \ x=41{}^\circ 4{9}'\ (\text{or})\ x=138{}^\circ 1{1}'\end{array}$

$ \displaystyle (\text{c})\ 3\ {{\sin }^{2}}x=4\sin x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 3{{\sin }^{2}}x-4\sin x=0\\\\\ \ \ \ \ \sin x(3\sin x-4)=0\\\\\ \ \ \ \ \sin x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{4}{3}\\\\(\text{i})\ \ \text{For }\sin x=0,\\\\\ \ \ \ \ \ x=0{}^\circ \ (\text{or})\ x=180{}^\circ \ (\text{or})\ x=360{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{4}{3}=1.333\\\ \\\ \ \ \ \ \text{Since }-1\le \sin x\le 1,\\\\\ \ \ \ \text{sin }x=\displaystyle \frac{4}{3}\text{ is impossible}\text{.}\end{array}$

$ \displaystyle (\text{d})\ 5\sin x\cos x=2\cos x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 5\sin x\cos x-2\cos x=0\\\\\ \ \ \ \ \cos x(5\sin x-2)=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{2}{5}\\\\(\text{i})\ \ \text{For }\cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{2}{5}=0.4\\\ \\\ \ \ \ x=23{}^\circ 3{5}'\ (\text{or})\ x=180{}^\circ -23{}^\circ 3{5}'\end{array}$

$ \displaystyle (\text{e})\ {{\cos }^{2}}x-\cos x=2$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ {{\cos }^{2}}x-\cos x-2=0\\\\\ \ \ \ \ (\cos x+1)(\cos x-2)=0\\\\\ \ \ \ \ \cos x=-1\ \ (\text{or})\ \cos x=2\\\\(\text{i})\ \ \text{For }\cos x=-1,\\\\\ \ \ \ \ \ x=180{}^\circ \ \\\\(\text{ii})\ \text{For }\cos x=2\\\ \\\ \ \ \ \ \text{Since }-1\le \cos x\le 1,\\\ \ \\\ \ \ \ \ \cos x=2\ \text{is not in domain}\text{.}\\\\\therefore \ \ \ x=180{}^\circ \ \text{is the only solution}\text{.}\end{array}$

$ \displaystyle (\text{f})\ 2\sin x\cos x-\cos x=0$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos x(2\sin x-1)=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{1}{2}\\\\(\text{i})\ \ \text{For }\cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{1}{2}\\\ \\\ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=180{}^\circ -30{}^\circ \\\\\ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=150{}^\circ \end{array}$

$ \displaystyle (\text{g})\ 2{{\sin }^{2}}x-\sin x=1$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 2{{\sin }^{2}}x-\sin x-1=0\\\\\ \ \ \ \ (2\sin x+1)(\sin x-1)=0\\\\\ \ \ \ \ \sin x=\displaystyle - \frac{1}{2}\ \ (\text{or})\ \sin x=1\\\\(\text{i})\ \ \text{For }\ \sin x=\displaystyle -\frac{1}{2},\\\\\ \ \ \ \ \ x=180{}^\circ +30{}^\circ \ (\text{or})\ x=360{}^\circ -30{}^\circ \\\\\ \ \ \ \ \ x=210{}^\circ \ (\text{or})\ x=330{}^\circ \\\\(\text{ii})\ \text{For }\sin x=1\\\ \\\ \ \ \ \ x=90{}^\circ \ \end{array}$

$ \displaystyle (\text{h})\ 2\sin x\cos x=\sqrt{3}\cos x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 2\sin x\cos x-\sqrt{3}\cos x=0\\\\\ \ \ \ \ \cos x(2\sin x-\sqrt{3})=0\\\\\ \ \ \ \ \cos x=0\ \ (\text{or})\ \sin x=\displaystyle \frac{{\sqrt{3}}}{2}\\\\(\text{i})\ \ \text{For }\ \cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ \ (\text{or})\ x=270{}^\circ \\\\(\text{ii})\ \text{For }\sin x=\displaystyle \frac{{\sqrt{3}}}{2}\\\ \\\ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=180{}^\circ -60{}^\circ \\\\\ \ \ \ \ x=60{}^\circ \ (\text{or})\ x=120{}^\circ \end{array}$

$ \displaystyle (\text{i})\ 2\sin x\cos x-\cos x+4\sin x-2=0$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \cos x(2\sin x-1)+2(2\sin x-1)=0\\\\\ \ \ \ \ (2\sin x-1)(\cos x+2)=0\\\\\ \ \ \ \ \sin x=\displaystyle \frac{1}{2}\ \ (\text{or})\ \cos x=-2\\\\(\text{i})\ \ \text{For }\ \sin x=\displaystyle \frac{1}{2},\\\\\ \ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=180{}^\circ -30{}^\circ \\\\\ \ \ \ \ \ x=30{}^\circ \ (\text{or})\ x=150{}^\circ \\\\(\text{ii})\ \text{For }\cos x=-2,\\\\\ \ \ \ \ \text{Since }-1\le \cos x\le 1,\ \\\\\ \ \ \ \ \cos x=-2\,\text{is}\ \text{out of domain and}\ \\\ \ \ \ \ \text{hence }\cos x=-2\ \text{has no solution}\text{.}\end{array}$

$ \displaystyle (\text{j})\ \ 8{{\cos }^{2}}x-2\cos x-5=\sec x$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 8{{\cos }^{2}}x-2\cos x-5=\displaystyle \frac{1}{{\cos x}}\\\\\ \ \ \ \ 8{{\cos }^{3}}x-2{{\cos }^{2}}x-5\cos x=1\\\\\ \ \ \ \ 8{{\cos }^{3}}x-2{{\cos }^{2}}x-5\cos x-1=0\\\\\ \ \ \ \ (2\cos x+1)(4\cos x+1)(x-1)=0\ \left[ {\text{using factor theorem}} \right]\\\\\therefore \ \ \ \cos x=\displaystyle -\frac{1}{2}\ (\text{or})\ \cos x=\displaystyle -\frac{1}{4}\ (\text{or})\ \cos x=1\\\\(\text{i})\ \ \text{For }\ \cos x=\displaystyle -\frac{1}{2},\\\\\ \ \ \ \ \ x=180{}^\circ -60{}^\circ (\text{or})\ x=180{}^\circ +60{}^\circ \\\\\ \ \ \ \ \ x=120{}^\circ \ (\text{or})\ x=240{}^\circ \\\\(\text{ii})\ \text{For }\cos x=\displaystyle -\frac{1}{4}=0.25,\\\\\ \ \ \ \ \ x=180{}^\circ -14{}^\circ 2{9}'\ (\text{or})\ x=180{}^\circ +14{}^\circ 2{9}'\\\\\ \ \ \ \ \ x=165{}^\circ 3{1}'\ (\text{or})\ x=194{}^\circ 2{9}'\\\\(\text{iii})\ \text{For }\ \cos x=0,\\\\\ \ \ \ \ \ x=90{}^\circ (\text{or})\ x=270{}^\circ \end{array}$

Exercise (11.3) - No(1) Sample Solutions

(b) Solution (1)

$ \displaystyle \sin \text{135}{}^\circ =\sin (\text{180}{}^\circ -45{}^\circ )=\sin 45{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 135{}^\circ \text{ =}\cos \text{(180}{}^\circ -45{}^\circ )=-\cos 45{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\tan 135{}^\circ \text{ =}-1\\\\\cot 135{}^\circ \text{ =}-1\\\\\sec 135{}^\circ \text{ =}-\sqrt{2}\\\\\operatorname{cosec}135{}^\circ \text{ =}\sqrt{2}\end{array}$

(b) Solution (2)

$ \displaystyle \begin{array}{l}\ \ \ \text{principal angle = 135}{}^\circ \\\\\therefore \text{basic acute angle = 45}{}^\circ \end{array}$

$ \displaystyle \therefore \ \sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}\operatorname{and}\ \cos \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \begin{array}{l}\ \ \ \sin \text{135}{}^\circ =\sin \text{45}{}^\circ \operatorname{and}\ \cos \text{135}{}^\circ =\cos \text{45}{}^\circ \ \text{numerically}\text{.}\\\\\ \ \ \text{But 135}{}^\circ \ \text{lies in the second quadrant}\text{.}\end{array}$

$ \displaystyle \therefore \sin \text{135}{}^\circ =\sin \text{45}{}^\circ =\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \cos \text{135}{}^\circ =-\cos \text{45}{}^\circ =-\frac{{\sqrt{2}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\cot {{{135}}^{{}^\circ }}\text{ =}-1} \\ {} \\ {\sec {{{135}}^{{}^\circ }}\text{ =}-\sqrt{2}} \\ {} \\ {\operatorname{cosec}{{{135}}^{{}^\circ }}\text{ =}\sqrt{2}} \end{array}$

(d) Solution (1)

$ \displaystyle \ \ \ \sin 210{}^\circ =\sin (180{}^\circ +30{}^\circ )=-\sin 30{}^\circ =\displaystyle -\frac{1}{2}$

$ \displaystyle \ \ \ \cos 210{}^\circ =\cos (180{}^\circ +30{}^\circ )=-\cos 30{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \tan 210{}^\circ =\displaystyle \frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \cot 210{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \sec 210{}^\circ =\displaystyle -\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \operatorname{cosec} 210{}^\circ =-2$

(d) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = 210}{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 30}{}^\circ $

$ \displaystyle \therefore \ \sin \text{30}{}^\circ =\frac{1}{2}\operatorname{and}\ \cos \text{30}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \sin \text{210}{}^\circ =\sin \text{30}{}^\circ \operatorname{and}\ \cos \text{210}{}^\circ =\cos \text{30}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But 210}{}^\circ \ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin \text{210}{}^\circ =\sin \text{30}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \cos \text{210}{}^\circ =-\cos \text{30}{}^\circ =\displaystyle -\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \begin{array}{*{20}{l}} {\tan \text{210}{}^\circ \text{ =}\displaystyle \frac{{\sqrt{3}}}{3}} \\ {} \\ {\cot \text{210}{}^\circ \text{ =}\sqrt{3}} \\ {} \\ {\sec \text{210}{}^\circ \text{ =}\displaystyle -\frac{{2\sqrt{3}}}{3}} \\ {} \\ {\operatorname{cosec}\text{210}{}^\circ \text{ =}-2} \end{array}$

(m) Solution (1)

$ \displaystyle \ \ \ \sin (-120{}^\circ )=-\sin 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos (-120{}^\circ )=\cos 120{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos (180{}^\circ -60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(m) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }-120{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin (-120{}^\circ )=\sin \text{60}{}^\circ \operatorname{and}\ \cos (-120{}^\circ)x =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But }(-120{}^\circ )\ \text{lies in the third quadrant}\text{.}$

$ \displaystyle \therefore \sin (-120{}^\circ )=\sin \text{60}{}^\circ =-\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos (-120{}^\circ )=-\cos \text{60}{}^\circ =-\frac{1}{2}$

$ \displaystyle \ \ \ \tan (-120{}^\circ )=\sqrt{3}$

$ \displaystyle \ \ \ \cot (-120{}^\circ )=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec (-120{}^\circ )=-2$

$ \displaystyle \ \ \ \operatorname{cosec}(-120{}^\circ )=-\frac{{2\sqrt{3}}}{3}$

(q) Solution (1)

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sin 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \ \cos 480{}^\circ =\cos (360{}^\circ +60{}^\circ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\cos 60{}^\circ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$

(q) Solution (2)

$ \displaystyle \ \ \ \text{principal angle = }480{}^\circ $

$ \displaystyle \therefore \text{basic acute angle = 60}{}^\circ $

$ \displaystyle \therefore \ \sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}\operatorname{and}\ \cos \text{60}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \sin 480{}^\circ =\sin \text{60}{}^\circ \operatorname{and}\ \cos 480{}^\circ =\cos \text{60}{}^\circ \ \text{numerically}\text{.}$

$ \displaystyle \ \ \ \text{But}\ 480{}^\circ \ \text{lies in the first quadrant}\text{.}$

$ \displaystyle \therefore \sin 480{}^\circ =\sin \text{60}{}^\circ =\frac{{\sqrt{3}}}{2}$

$ \displaystyle \ \ \ \cos 480{}^\circ =\cos \text{30}{}^\circ =\frac{1}{2}$

$ \displaystyle \ \ \ \tan 480{}^\circ =\sqrt{3}$

$ \displaystyle \ \ \ \cot 480{}^\circ =\frac{{\sqrt{3}}}{3}$

$ \displaystyle \ \ \ \sec 480{}^\circ =2$

$ \displaystyle \ \ \ \operatorname{cosec}480{}^\circ =\frac{{2\sqrt{3}}}{3}$

Trigonometric Ratios of 0°, 180°, 270° and 360°

Trigonometric Ratios of 0° 

From the unit circle we have ,

$ \displaystyle \sin 0°=y=0$

$ \displaystyle \cos 0°=x=1$

Therefore

$ \displaystyle \tan 0°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 0°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 0°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 0°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 90° 

Similarly,

$ \displaystyle \sin 90°=y=1$

$ \displaystyle \cos 90°=x=0$

$ \displaystyle \tan 90°=\frac{y}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \cot 90°=\frac{x}{y}=\frac{0}{1}=0$

$ \displaystyle \sec 90°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 90°=\frac{1}{y}=\frac{1}{1}=1$

Trigonometric Ratios of 180° 

Similarly,

$ \displaystyle \sin 180°=y=0$

$ \displaystyle \cos 180°=x=-1$

$ \displaystyle \tan 180°=\frac{y}{x}=\frac{0}{-1}=0$

$ \displaystyle \cot 180°=\frac{x}{y}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \sec 180°=\frac{1}{x}=\frac{1}{-1}=-1$

$ \displaystyle \operatorname{cosec} 180°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$

Trigonometric Ratios of 270° 

Similarly,

$ \displaystyle \sin 270°=y=-1$

$ \displaystyle \cos 270°=x=0$

$ \displaystyle \tan 270°=\frac{y}{x}=\frac{-1}{0}=\text{undefined}$

$ \displaystyle \cot 270°=\frac{x}{y}=\frac{0}{-1}=0$

$ \displaystyle \sec 270°=\frac{1}{x}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \operatorname{cosec} 270°=\frac{1}{y}=\frac{1}{-1}=-1$

Trigonometric Ratios of 360° 

Similarly,

$ \displaystyle \sin 360°=y=0$

$ \displaystyle \cos 360°=x=1$

$ \displaystyle \tan 360°=\frac{y}{x}=\frac{0}{1}=1$

$ \displaystyle \cot 360°=\frac{x}{y}=\frac{1}{0}=\text{undefined}$

$ \displaystyle \sec 360°=\frac{1}{x}=\frac{1}{1}=1$

$ \displaystyle \operatorname{cosec} 360°=\frac{1}{y}=\frac{1}{0}=\text{undefined}$


Applet တြင္ undefined အတြက္ ∞ သေကၤတ ကို သံုးထားပါသည္။

Trigonometric Ratios of Special Angles

အနား တစ္ဖက္ $ \displaystyle x$ unit ရွိတဲ့ စတုရနး္ $ \displaystyle ABCD$ ဆိုပါစို႔။

စတုရန္း ျဖစ္ေသာေၾကာင့္ ေထာင့္ျဖတ္မ်ဥ္း $ \displaystyle AC$ က သက္ဆိုင္ရာ ေထာင့္မ်ားကို ထက္၀က္ပိုင္းပါတယ္။ ဒါဆိုရင္ရင္ ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္း ထပ္တူညီ ေထာင့္မွန္ ႀတိဂံႏွစ္ခု ျဖစ္လာပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ ႏွစ္နားညီ တဲ့ ေတာင့္မွန္ ႀတိဂံတစ္ခု  ($ \displaystyle 45°-45°$ right triangle လို႔ ေခၚပါတယ္) ရလာပါတယ္။ $ \displaystyle AB=BC=x$ ျဖစ္တာ ေၾကာင့္ $ \displaystyle AC$ ရဲ့ အလ်ားကို Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

        $ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ =2{{x}^{2}}\\\\\therefore \ \ AC=\sqrt{2}x\end{array}$

ဒါဆိုရင္ $ \displaystyle AB:BC:AC=1:1:\sqrt{2}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 45°-45°$ right triangle ကို special triangle လို႔ ေခၚၿပီး $ \displaystyle 45°$ ေထာင့္ကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

အနားေတြရဲ့ အခ်ိဳးကို သိၿပီဆိုေတာ့ $ \displaystyle 45°$ angle ရဲ့ trigonometric ratios ေတြကို အလြယ္ တကူရွာႏိုင္ၿပီေပါ့။

Trigonometric Ratios of $ \displaystyle 45°$

$ \displaystyle \sin 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \cos 45{}^\circ =\frac{x}{{\sqrt{2}x}}=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}$

$ \displaystyle \tan 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \cot 45{}^\circ =\frac{x}{x}=1$

$ \displaystyle \sec 45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

$ \displaystyle \operatorname{cosec}45{}^\circ =\frac{{\sqrt{2}x}}{x}=\sqrt{2}$

ယခုတစ္ခါ အနားတစ္ဘက္ $ \displaystyle 2x$ unit ရွိတဲ့ သံုးနားညီႀတိဂံတစ္ခု $ \displaystyle ABD$ ကို စဥ္းစားၾကည့္မယ္။ ေထာင့္စြန္းမွတ္ $ \displaystyle A$ မွ $ \displaystyle BD$ ေပၚသို႔ အျမင့္မ်ဥ္း $ \displaystyle AC$ ကိုဆြဲလိုက္မယ္ ဆိုရင္ သံုးနားညီ ႀတိဂံျဖစ္တာေၾကာင့္ $ \displaystyle AC$ ဟာ အလယ္မ်ဥ္းလည္း ျဖစ္သလို ေထာင့္ထက္၀က္ပိုင္း မ်ဥ္းလဲ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္ ပံုမွာ ျပထားတဲ့ အတိုင္း $ \displaystyle AC$ ဟာ $ \displaystyle \vartriangle ABD$ ကို ထပ္တူညီ ႀတိဂံ ႏွစ္ခုအျဖစ္ ပိုင္းျဖတ္ လိုက္ပါတယ္။

၎တို႔အထဲက ေထာင့္မွန္ႀတိဂံ $ \displaystyle ABC$ ကို ခြဲထုတ္လိုက္ရင္ $ \displaystyle \vartriangle ABC$ ဟာ $ \displaystyle 30°-60°$ right triangle ျဖစ္ၿပီး $ \displaystyle AC$ ရဲ့ အလ်ားကိုေတာ့ Pythagoras' Theorem နဲ႔ တြက္ယူႏိုင္ပါတယ္။

 Pythagoras' Theorem အရ

$ \displaystyle \begin{array}{l}\ \ \ \ A{{C}^{2}}=A{{B}^{2}}-B{{C}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}-{{x}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ =3{{x}^{2}}\\\\\therefore \ \ \ \ AC=\sqrt{3}x\end{array}$

ဒါဆိုရင္ $ \displaystyle BC:AB:AC=1:2:\sqrt{3}$  ဆိုၿပီး အနားအခ်ိဳးေတြကို အလြယ္တကူ သိႏိုင္ပါၿပီ။ လက္ေတြ႕ တိုင္းတာမႈ မလုပ္ပဲ သခၤ်ာရဲ့ မွန္ကန္ခ်က္မ်ားျဖင့္ အနားအခ်ိဳးေတြကို အလြယ္တကူရွာႏိုင္တဲ့ $ \displaystyle 30°-60°$ right triangle ကိုလည္း special triangle လို႔ ေခၚၿပီး $ \displaystyle 30°$ နဲ႔ $ \displaystyle 60°$ ေထာင့္ ေတြကိုေတာ့ special angle လို႔ ေခၚပါတယ္။

Trigonometric Ratios of $ \displaystyle 30°$

$ \displaystyle \sin 30{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \cos 30{}^\circ =\frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \tan 30{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \cot 30{}^\circ =\frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \sec 30{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$

$ \displaystyle \operatorname{cosec}30{}^\circ =\frac{{2x}}{x}=2$

Trigonometric Ratios of $ \displaystyle 60°$

$ \displaystyle \sin 60{}^\circ = \frac{{\sqrt{3}x}}{{2x}}=\frac{{\sqrt{3}}}{2}$

$ \displaystyle \cos 60{}^\circ =\frac{x}{{2x}}=\frac{1}{2}$

$ \displaystyle \tan 60{}^\circ = \frac{{\sqrt{3}x}}{x}=\sqrt{3}$

$ \displaystyle \cot 60{}^\circ =\frac{x}{{\sqrt{3}x}}=\frac{1}{{\sqrt{3}}}=\frac{{\sqrt{3}}}{3}$

$ \displaystyle \sec 60{}^\circ = \frac{{2x}}{x}=2$

$ \displaystyle \operatorname{cosec}60{}^\circ =\frac{{2x}}{{\sqrt{3}x}}=\frac{2}{{\sqrt{3}}}=\frac{{2\sqrt{3}}}{3}$