Question 4 and 5, Exercise 10.2

Solutions of Question 4 and 5 of Exercise 10.2 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\cos \theta =-\dfrac{3}{7}$ and terminal ray of $\theta$ is in the third quadrant, then find $\sin \dfrac{\theta }{2}$.

Given: $\cos \theta =-\dfrac{3}{7}$ and terminal ray of $\theta$ is in the third quadrant, that is,

$$\begin{align}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies &\frac{\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{4}\end{align}$$ This gives $\frac{\theta}{2}$ lies in 2nd quadrant.

By using the half angle identity: $$\sin\dfrac{\theta}{2}=\pm\sqrt{\dfrac{1-\cos \theta }{2}}$$ As $\frac{\theta}{2}$ lies in 2nd quadrant and $\sin$ is positive in 2nd quadrant, therefore $$\begin{align}\sin\dfrac{\theta }{2}&=-\sqrt{\dfrac{1-\cos \theta }{2}}\\ &=-\sqrt{\dfrac{1-\left( -\dfrac{3}{7} \right)}{2}}\\ &=-\sqrt{\dfrac{1+\dfrac{3}{7}}{2}}\end{align}$$ $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{\theta}{2}=-\sqrt{\dfrac{5}{7}}}$$

Use double angle identities to evaluate exactly $\sin \dfrac{2\pi }{3}$.

Given: $\sin \dfrac{2\pi }{3}$.
By using double angle identities, we have $$\begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ \implies \sin 2\left(\dfrac{\pi }{3}\right)&=2\sin \dfrac{\pi }{3}\cos \dfrac{\pi}{3}\\ &=2\left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{1}{2} \right)\end{align}$$ $$\implies \bbox[4px,border:2px solid black]{\sin\dfrac{2\pi}{3}={\dfrac{\sqrt{3}}{2}}}$$

Use the double angle identities to evaluate exactly $\cos \dfrac{2\pi }{3}$.

Given: $\cos \dfrac{2\pi }{3}$
By using double angle identities, we have $$\begin{align}\cos 2\theta &=2{{\cos }^{2}}\theta -1\\ \implies \cos 2\left(\dfrac{\pi }{3}\right)&=2{{\cos }^{2}}\dfrac{\pi }{3}-1\\ &=2{{\left( \dfrac{1}{2} \right)}^{2}}-1\end{align}$$ $$\implies \bbox[4px,border:2px solid black]{\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}}$$

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