Question
Evaluate the following:
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$

Answer

We have to find the value of the following expression
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ\dots(1)$
$\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2},\sin60^\circ=\frac{\sqrt3}{2}\cos60^\circ=\frac{1}{2}$
Now
So by substituting above values in equation (1)
We get,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
$=\frac{1}{2}\times\frac{1}{\sqrt2}-\frac{\sqrt3}{2}\times\frac{1}{\sqrt2}$
$=\frac{1}{2\sqrt2}-\frac{\sqrt3}{2\sqrt2}$
$=\frac{1-\sqrt3}{2\sqrt2}$
Therefore,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ=\frac{1-\sqrt3}{2\sqrt2}$

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