Problems Study

 

1. Functions f and g are such that  g^-1 (x) = x  -${\displaystyle \frac{1}{3 }}$ and (f ∘ g ) (x) = 3x - 1.
    Find (g ^-1 ∘ f ) (x) where x ∈ R.

   Solution 

   Let g ^-1(x ) = y then g (y) = x.
   Hence x - ${\displaystyle \frac{1}{3}}$ = y
            x = y + ${\displaystyle \frac{1}{3}}$
            g (y) = y + ${\displaystyle \frac{1}{3}}$
            g (x) = x + ${\displaystyle \frac{1}{3}}$
           (f ∘ g) (x) = 3x - 1
           f (g ) (x)   = 3x - 1
           f (x + ${\displaystyle \frac{1}{3}}$)  = 3x - 1
                          = 3 (x + ${\displaystyle \frac{1}{3}}$) - 2
   Hence f (x ) = 3x - 2
           (g ^-1 ∘ f ) (x) = g ^-1 ( f   (x ) )
                             = g ^-1 ( 3x - 2)
                             = 3x - ${\displaystyle \frac{7}{3}}$

2. If x, y and z are any three consecutive even numbers, Show that x² + y² + z² = 3y² + 8.

   Solution 

   Let x = 2a where a is an integer.

   Since x, y and z are any three consecutive even numbers,

   y = 2a + 2 and z = 2a + 4

   x² + y² + z² = (2a)² + (2a + 2)² + (2a + 4)² 

                     = 4a² + 4a² + 8a + 4 + 4a² + 16a + 16

                     = 12a² + 24a + 20

                     = 12a² + 24a + 12 + 8

                     = 3 (4a² + 8a + 4) + 8 

                     = 3 (2a + 2)² + 8 

                     = 3 y² + 8

2012 ပထမလပတ္ စာေမးပြဲ ေမးခြန္းနဲ႔ အေျဖ

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Grade (11)



Grade (10)

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}.$

Show/Hide Solution

(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}$