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)$

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$ \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 (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}$