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Question

Find the exact value of the given expression.

sin15cos15\sin 15^{\circ} \cos 15^{\circ}.

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Answered 2 years ago
Answered 2 years ago
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sin15°cos15°=12(2sin15°cos15°)=(1)12sin(215°)=12sin30°=1212=14\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)sin2α=2sinαcosα\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)\\\\ 3)\tan 2\alpha=2tanα1tan2α4)\dfrac{2\tan \alpha}{1-\tan^2\alpha}\\\\ 4)\sinα2\dfrac{\alpha}{2}=\pm1cosα25)\sqrt{\dfrac{1-\cos\alpha}{2}}\\\\ 5)\cosα2\dfrac{\alpha}{2}=\pm1+cosα26)\sqrt{\dfrac{1+\cos\alpha}{2}}\\\\ 6)\tanα2\dfrac{\alpha}{2}=\pm1cosα1+cosα\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}=sinα1+cosα\dfrac{\sin\alpha}{1+\cos\alpha}=$$\dfrac{1-\cos\alpha}{\sin\alpha}$

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