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*}
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*}
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*}
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*}
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