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.

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 $\theta = 15^{\circ}$ $$\begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}\sin 2(15^{\circ}) \\ & \frac{1}{2}\sin 30^{\circ} = \frac{1}{2} \times \frac{1}{2} \end{align*}$$ $$\begin{align*} \implies \boxed{\sin 15^{\circ} \cos 15^{\circ} = \frac{1}{4}} \end{align*}$$

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 $\theta = 15^{\circ}$ $$\begin{align*} \cos^2 15^\circ - \sin^2 15^\circ & = \cos 2(15^\circ) \\ & = \cos 30^\circ \\ & = \frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \implies \boxed{\cos^2 15^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{2}} \end{align*}$$

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 $$\begin{align*} 1-2\sin^2 \left(\frac{\pi}{8}\right)&=\cos 2\left(\frac{\pi}{8}\right)\\ &=\cos \left(\frac{\pi}{4}\right)\\ \end{align*}$$ $$\begin{align*} \implies \boxed{1-2\sin^2 \left(\frac{\pi}{8}\right) = \frac{1}{\sqrt{2}}} \end{align*}$$

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 $$\begin{align*} 2\cos^2 \left(\frac{\pi}{12}\right)-1&=\cos 2\left(\frac{\pi}{12}\right)\\ &=\cos \left(\frac{\pi}{6}\right)\\ \end{align*}$$ $$\begin{align*} \implies \boxed{2\cos^2 \left(\frac{\pi}{12}\right)-1 = \frac{1}{2}} \end{align*}$$

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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