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- Question 13, Exercise 10.1
- &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ \implies &r^2\left(\cos^2\varphi+\sin^2\varphi\right) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{
- Question11 and 12, Exercise 10.1
- \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\alpha +\beta =180^\circ-\gamma \\ \implies &\dfrac{\alpha +\beta }{2}=\dfrac{180^\circ-\gamma }{2}\\ \implies &\dfrac{\alpha}{2}+\dfrac{\beta}{2}=90^\circ-\dfr... ght)=\tan \left( 90-\dfrac{\gamma }{2} \right)\\ \implies &\dfrac{\tan\dfrac{\alpha}{2}+\tan\dfrac{\beta}{2
- Question 5, Exercise 10.1
- dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\be... ht)\\ &=-\frac{3}{13}+\frac{48}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin(\alpha +\b... dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\be... t)\\ &=-\frac{20}{65}+\frac{36}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos(\alpha +\b
- Question 2, Exercise 10.2
- right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the ... right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\... right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the ... &=\dfrac{144}{169}-\dfrac{25}{169}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos 2\theta=-\
- Question 4 and 5, Exercise 10.2
- \begin{align}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies &\frac{\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{... =-\sqrt{\dfrac{1+\dfrac{3}{7}}{2}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{\the... lign}\sin 2\theta &=2\sin \theta \cos \theta \\ \implies \sin 2\left(\dfrac{\pi }{3}\right)&=2\sin \dfrac{... \right)\left( \dfrac{1}{2} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{2\pi
- Question 1, Exercise 10.2
- 26}} \right)\\ &=-\dfrac{10}{26} \end{align} $$ \implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\... {26}-\dfrac{1}{26}=\dfrac{24}{26}\end{align} $$ \implies \bbox[4px,border:2px solid black]{\cos2\theta=\df... rac{\frac{-5}{13}}{\frac{12}{13}}\end{align} $$ \implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\
- Question 1, Exercise 10.3
- \ &=\sin {{178}^{\circ }}-\sin {{68}^{\circ }}\\ \implies \sin {{55}^{\circ }}\cos {{123}^{\circ }}&=\dfrac... {A-B}{2} \right)\\ &=\sin A+\sin B.\end{align} $$\implies \sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}=\dfrac{1}{... c{P-Q}{2}\right)\\ &=\cos P-\cos Q \end{align} $$\implies\sin \dfrac{P+Q}{2}\cos \dfrac{P-Q}{2}=\dfrac{1}{2
- Question 3, Exercise 10.2
- right)\left( -\dfrac{3}{5} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\... rac{3}{5}}{2}}=\sqrt{\dfrac{2}{10}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos \dfrac{\th
- Question 8 and 9, Exercise 10.2
- cos 2\theta +\cos 4\theta \right] \end{align} $$\implies \bbox[4px,border:2px solid black]{\cos^4 \theta=\