# How do you find the exact value of sin60°cos30° + sin30°cos60°?

Apr 26, 2015

${30}^{o} \text{ and " 60^o "are angles of one of the standard triangles}$

$\sin \left({30}^{o}\right) = \frac{1}{2}$

$\cos \left({30}^{o}\right) = \frac{\sqrt{3}}{2}$

$\sin \left({60}^{o}\right) = \frac{\sqrt{3}}{2}$

$\cos \left({60}^{o}\right) = \frac{1}{2}$

So
$\sin \left({60}^{o}\right) \cdot \cos \left({30}^{o}\right) + \sin \left({30}^{o}\right) \cdot \cos \left({60}^{o}\right)$

$= \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4}$

$= 1$