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.

Use the product-to-sum formula to change the sum or difference: $\sin(-u) \sin 5u$.

Solution.

$$\begin{align*} &\sin(-u) \sin 5u \\ =& -\sin u \sin 5u \\ =& -\frac{1}{2}[\cos(u - 5u) - \cos(u + 5u)] \\ = &-\frac{1}{2}[\cos(-4u) - \cos(6u)] \\ =& \frac{1}{2}[\cos(6u) - \cos(4u) ] \end{align*}$$

Use the product-to-sum formula to change the sum or difference: $-2 \sin 100^{\circ}\sin (-20^{\circ})$.

Solution.

$$\begin{align*} &-2 \sin 100^{\circ} \sin (-20^{\circ}) \\ &= 2 \sin 100^{\circ} \sin 20^{\circ} \\ &= \cos(100^{\circ} - 20^{\circ}) - \cos(100^{\circ} + 20^{\circ}) \\ &= \cos 80^{\circ} - \cos 120^{\circ} \end{align*}$$

Use the product-to-sum formula to change the sum or difference: $\cos 23^{\circ} \sin 17^{\circ}$.

Solution.

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

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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