====== Question 3, Exercise 10.2 ======
Solutions of Question 3 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.
=====Question 3(i)=====
If $\sin \theta =\dfrac{4}{5}$ and terminal ray of $\theta$ is in the second quadrant, then find $\sin2\theta$.
====Solution====
Given: $\sin \theta =\dfrac{4}{5}$

Terminal ray of $\theta$ is in the second quadrant  and by drawing the reference triangle as shown:

{{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}}

We find: $\cos \theta =-\dfrac{3}{5}$.

Thus, we have the following by using double angle identity:
\begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\
&=2\left( \dfrac{4}{5} \right)\left( -\dfrac{3}{5} \right)\end{align} 
$$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\dfrac{24}{25}.}$$

=====Question 3(ii)=====
If $\sin \theta =\dfrac{4}{5}$ and terminal ray of $\theta$ is in the second quadrant, then find $\cos \dfrac{\theta }{2}$.
====Solution====
Given: $\sin \theta =\dfrac{4}{5}$

Terminal ray of $\theta$ is in the second quadrant  and by drawing the reference triangle as shown:

{{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}}

We find: $\cos \theta =-\dfrac{3}{5}$. 

Thus, we have the following by using half angle identities:
\begin{align}\cos \dfrac{\theta }{2}&=\sqrt{\dfrac{1+\cos \theta }{2}}\\
&=\sqrt{\dfrac{1-\dfrac{3}{5}}{2}}=\sqrt{\dfrac{2}{10}}\end{align}
$$\implies \bbox[4px,border:2px solid black]{\cos \dfrac{\theta }{2}=\dfrac{1}{\sqrt{5}}}$$

====Go to====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit10:ex10-2-p2|< Question 2]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit10:ex10-2-p4|Question 4 >]]</btn></text>