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.

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*}

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*}

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*}

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*}

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*}

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