Question 6 Exercise 8.2
Solutions of Question 6 of Exercise 8.2 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 6(i)
Use a double-angle identity to find exact values for the expression: $\sin 15^{\circ} \cos 15^{\circ}$
Solution.
We have double-angle identity:
$$\sin 2 \theta = 2\sin\theta \cos\theta$$
This gives
$$\sin\theta \cos\theta = \frac{1}{2}\sin 2\theta$$
Put $\theta = 15^{\circ}$
\begin{align*}
\sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}\sin 2(15^{\circ}) \\
& \frac{1}{2}\sin 30^{\circ} = \frac{1}{2} \times \frac{1}{2}
\end{align*}
\begin{align*}
\implies \boxed{\sin 15^{\circ} \cos 15^{\circ} = \frac{1}{4}}
\end{align*}
Question 6(ii)
Use a double-angle identity to find exact values for the expressions: $\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}$
Solution.
We have double-angle identity: $$\cos^2\theta -\sin^2\theta= \cos 2\theta = $$
Put $\theta = 15^{\circ}$ \begin{align*} \cos^2 15^\circ - \sin^2 15^\circ & = \cos 2(15^\circ) \\ & = \cos 30^\circ \\ & = \frac{\sqrt{3}}{2} \end{align*} \begin{align*} \implies \boxed{\cos^2 15^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{2}} \end{align*}
Question 6(iii)
Use a double-angle identity to find exact values for the expression: $1-2 \sin ^{2}\left(\frac{\pi}{8}\right)$
Solution.
We have a double-angle identity:
$$\cos 2\alpha = 1-2\sin^2 \alpha.$$
That is
$$1-2\sin^2 \alpha=\cos 2\alpha.$$
Put $\alpha= \dfrac{\pi}{8}$, we have
\begin{align*}
1-2\sin^2 \left(\frac{\pi}{8}\right)&=\cos 2\left(\frac{\pi}{8}\right)\\
&=\cos \left(\frac{\pi}{4}\right)\\
\end{align*}
\begin{align*}
\implies \boxed{1-2\sin^2 \left(\frac{\pi}{8}\right) = \frac{1}{\sqrt{2}}}
\end{align*}
Question 6(iv)
Use a double-angle identity to find exact values for the expression: $2 \cos ^{2}\left(\frac{\pi}{12}\right)-1$
Solution.
We have a double-angle identity: $$\cos 2\alpha = 2\cos^2 \alpha -1.$$ That is $$2\cos^2 \alpha -1=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{12}$, we have \begin{align*} 2\cos^2 \left(\frac{\pi}{12}\right)-1&=\cos 2\left(\frac{\pi}{12}\right)\\ &=\cos \left(\frac{\pi}{6}\right)\\ \end{align*} \begin{align*} \implies \boxed{2\cos^2 \left(\frac{\pi}{12}\right)-1 = \frac{1}{2}} \end{align*}
Question 6(v)
Use a double-angle identity to find exact values for the expression: $\frac{2 \tan \left(\dfrac{\pi}{12}\right)}{1-\tan ^{2}\left(\dfrac{\pi}{12}\right)}$
Solution.
We have a double-angle identity: $$\tan 2\alpha = \dfrac{2\tan\alpha}{1-2\tan\alpha}$$ Put \(\theta = \frac{\pi}{12}\): \begin{align*} \frac{2 \tan \left(\dfrac{\pi}{12}\right)}{1-\tan ^{2}\left(\dfrac{\pi}{12}\right)} & = \tan 2\left(\dfrac{\pi}{12}\right) \\ & = \tan \left(\dfrac{\pi}{6}\right) \\ & = \frac{1}{\sqrt{3}} \end{align*} \begin{align*} \implies \boxed{\frac{2 \tan \left(\dfrac{\pi}{12}\right)}{1-\tan ^{2}\left(\dfrac{\pi}{12}\right)} = \frac{1}{\sqrt{3}}} \end{align*}
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