## Related questions with answers

Find the exact value of the given expression.

$\sin 15^{\circ} \cos 15^{\circ}$.

Solution

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$\begin{align} \sin 15\text{\textdegree}\cos 15\text{\textdegree}&=\dfrac{1}{2}(2\sin 15\text{\textdegree}\cos 15\text{\textdegree}) \\\\ &\overset{(1)}{=}\dfrac{1}{2}\sin(2\cdot 15\text{\textdegree}) \\\\ &=\dfrac{1}{2}\cdot \sin 30\text{\textdegree}\\\\ &=\dfrac{1}{2}\cdot\dfrac{1}{2}\\\\ &=\dfrac{1}{4} \end{align}$

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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