Question

Evaluate the following :
cos ${60}^{\circ }$ cos ${45}^{\circ }$ - sin ${60}^{\circ }$ sin ${45}^{\circ }$

Answer 1

Given $\cos {60}^{\circ }\cos {45}^{\circ }-\sin {60}^{\circ }\sin {45}^{\circ }$

$=\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)$ $[\because \sin {45}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}},\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\cos {60}^{\circ }=\frac{1}{2}]$

$=\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}$

$=\frac{(1-\sqrt{3})}{2\sqrt{2}}$