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Question

Find the exact value of the given expression.$1-2 \sin ^{2} \frac{7 \pi}{12}$.

Solution

Verified
Step 1
1 of 2

Work:

\begin{align} 1-2\sin^2\dfrac{7\pi}{12}&\overset{(2)}{=} \cos(2\cdot \dfrac{7\pi}{12}) \\\\ &=\cos \dfrac{7\pi}{6} \\\\ &=-\dfrac{\sqrt 3}{2} \end{align}

Double-Angle and Half-Angle Formulas:$\\\\$ 1)$\sin 2\alpha=2\sin \alpha\cos \alpha$\\ 2)$\cos 2\alpha=\cos^2\alpha-\sin^2\alpha= =1-2\sin^2\alpha=2\cos^2\alpha-1 $\\\\ 3)$\tan 2\alpha=$\dfrac{2\tan \alpha}{1-\tan^2\alpha}\\\\ 4)$\sin$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1-\cos\alpha}{2}}\\\\ 5)$\cos$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1+\cos\alpha}{2}}\\\\ 6)$\tan$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}$=$\dfrac{\sin\alpha}{1+\cos\alpha}$=$$\dfrac{1-\cos\alpha}{\sin\alpha}$

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