====== Question 2(i, ii, iii, iv and v) Exercise 8.3 ======
Solutions of Question 2(i, ii, iii, iv and v) of Exercise 8.3 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 2(i)=====
Rewrite the sum or difference as a product of two function: $\sin 70^{\circ} + \sin 30^{\circ}$

** Solution. **

\begin{align*}
 & \quad \sin 70^{\circ} + \sin 30^{\circ} \\
& = 2 \sin \left(\frac{70+30}{2} \right) \cos \left(\frac{70-30}{2} \right) \\
& = 2 \sin \left(\frac{100}{2} \right) \cos \left(\frac{40}{2} \right) \\
& = 2 \sin 50^\circ \cos 20^\circ 
\end{align*}
GOOD


=====Question 2(ii)=====
Rewrite the sum or difference as a product of two function: $\sin 76^{\circ} - \sin 14^{\circ}$

** Solution. **

\begin{align*}
& \quad \sin 76^{\circ} - \sin 14^{\circ} \\
& = 2 \cos \left(\frac{76 + 14}{2}\right) \sin \left(\frac{76 - 14}{2}\right) \\
& = 2 \cos \left(\frac{90}{2}\right) \sin \left(\frac{62}{2}\right) \\
& = 2 \cos 45^\circ \sin 31^\circ \\
\end{align*}

=====Question 2(iii)=====
Rewrite the sum or difference as a product of two function: $\cos 58^{\circ} + \cos 12^{\circ}$

** Solution. **

\begin{align*}
&\quad  \cos 58^{\circ} + \cos 12^{\circ} \\
& = 2 \cos \left(\frac{58 + 12}{2}\right) \cos \left(\frac{58 - 12}{2}\right) \\
& = 2 \cos \left(\frac{70}{2}\right) \cos \left(\frac{46}{2}\right) \\
& = 2 \cos 35^\circ \cos 23^\circ 
\end{align*}

=====Question 2(iv)=====
Rewrite the sum or difference as a product of two function: $\cos \frac{p-q}{2} + \cos \frac{p+q}{2}$

** Solution. **

\begin{align*}
&\quad  \cos \frac{p-q}{2} + \cos \frac{p+q}{2} \\
& = 2 \cos \left( \frac{(p-q) + (p+q)}{2} \right) \cos \left( \frac{(p-q) - (p+q)}{2} \right) \\
& = 2 \cos \left( \frac{2p}{2} \right) \cos \left( \frac{-2q}{2} \right) \\
& = 2 \cos p \cos (-q) \\
& = 2 \cos p \cos q
\end{align*}

=====Question 2(v)=====
Rewrite the sum or difference as a product of two function: $\sin (-10^{\circ}) + \sin (-20^{\circ})$

** Solution. **

\begin{align*}
&\quad  \sin (-10^{\circ}) + \sin (-20^{\circ}) \\
& = -\sin 10^{\circ} - \sin 20^{\circ} \\
& = -\left( \sin 10^{\circ} + \sin 20^{\circ} \right) \\
& = -2 \sin \left( \frac{10^{\circ} + 20^{\circ}}{2} \right) \cos \left( \frac{10^{\circ} - 20^{\circ}}{2} \right) \\
& = -2 \sin \left( \frac{30^{\circ}}{2} \right) \cos \left( \frac{-10^{\circ}}{2} \right) \\
& = -2 \sin 15^{\circ} \cos (-5^{\circ}) \\
& = -2 \sin 15^{\circ} \cos 5^{\circ}
\end{align*}
GOOD

====Go to ====

<text align="left"><btn type="primary">[[math-11-nbf:sol:unit08:ex8-3-p3|< Question 1(ix, x & xi)  ]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit08:ex8-3-p5|Question 3(i, ii, iii, iv & v) >]]</btn></text>