# How do you find the exact functional value sin(60˚+45˚) using the cosine sum or difference identity?

Aug 14, 2015

$\sin \left({60}^{\circ} + {45}^{\circ}\right) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

#### Explanation:

Using the sine identity:
$\sin \left(A \pm B\right) = \sin A \cos B \pm \cos A \sin B$

$\sin \left({60}^{\circ} + {45}^{\circ}\right) = \sin {60}^{\circ} \cos {45}^{\circ} + \cos {60}^{\circ} \sin {45}^{\circ}$
$= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

If you want to use the cosine identity:
 cos (A +- B) = cos A cos B ""_+^(-) sin A sin B

$\sin A = \cos \left(A - {90}^{\circ}\right)$

$\sin \left({60}^{\circ} + {45}^{\circ}\right) = \cos \left({60}^{\circ} - {45}^{\circ}\right)$
$= \cos {60}^{\circ} \cos {45}^{\circ} + \sin {60}^{\circ} \sin {45}^{\circ}$
$= \frac{1}{2} \times \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}$
$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$