====== Question 1(v, vi, vii & viii) Exercise 8.3 ======
Solutions of Question 1(v, vi, vii & viii) 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 1(v)=====
Use the product-to-sum formula to change the sum or difference: $ \sin(-u) \sin 5u$.

** Solution. **

\begin{align*}
&\sin(-u) \sin 5u \\
=& -\sin u \sin 5u \\
=& -\frac{1}{2}[\cos(u - 5u) - \cos(u + 5u)] \\
= &-\frac{1}{2}[\cos(-4u) - \cos(6u)] \\
=& \frac{1}{2}[\cos(6u) - \cos(4u) ]
\end{align*}
GOOD
=====Question 1(vi)=====
Use the product-to-sum formula to change the sum or difference: $-2 \sin 100^{\circ}\sin (-20^{\circ}) $.

** Solution. **

\begin{align*}
&-2 \sin 100^{\circ} \sin (-20^{\circ}) \\
&= 2 \sin 100^{\circ} \sin 20^{\circ} \\
&= \cos(100^{\circ} - 20^{\circ}) - \cos(100^{\circ} + 20^{\circ}) \\
&= \cos 80^{\circ} - \cos 120^{\circ}
\end{align*}
GOOD :!:
=====Question 1(vii)=====
Use the product-to-sum formula to change the sum or difference: $\cos 23^{\circ} \sin 17^{\circ}$.

** Solution. **

\begin{align*}
&\cos 23^{\circ} \sin 17^{\circ} \\
=& \frac{1}{2}[ \sin(23^{\circ} + 17^{\circ}) - \sin(23^{\circ} - 17^{\circ}) ] \\
= &\frac{1}{2}[ \sin 40^{\circ} - \sin 6^{\circ} ]
\end{align*}
GOOD

=====Question 1(viii)=====
Use the product-to-sum formula to change the sum or difference: $2 \cos56^{\circ} \sin48^{\circ}$.

** Solution. **

\begin{align*}
&2 \cos 56^{\circ} \sin 48^{\circ} \\
=& [ \sin(56^{\circ} + 48^{\circ}) - \sin(56^{\circ} - 48^{\circ}) ] \\
=&  \sin 104^{\circ} - \sin 8^{\circ}
\end{align*}
GOOD



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