Evaluate each of the following:$ \quad\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right) $

Given:

\( \quad\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right) \)

To do:

We have to evaluate \( \quad\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right) \).

Solution:  

We know that,

$\cos 0^{\circ}=1$

$\sin 45^{\circ}=\frac{1}{\sqrt2}$

$\sin 30^{\circ}=\frac{1}{2}$

$\sin 90^{\circ}=1$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

$\cos 60^{\circ}=\frac{1}{2}$

Therefore,

$\quad\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right)=\left[ 1+\left(\frac{1}{\sqrt{2}}\right) +\left(\frac{1}{2}\right)\right]\left[ 1-\left(\frac{1}{\sqrt{2}}\right) +\left(\frac{1}{2}\right)\right]$

$=\left[\frac{1\left( 2\sqrt{2}\right) +1( 2) +1\left(\sqrt{2}\right)}{2\sqrt{2}}\right]\left[\frac{1\left( 2\sqrt{2}\right) -1( 2) +1\left(\sqrt{2}\right)}{2\sqrt{2}}\right]$

$=\left[\frac{3\sqrt{2} +2}{2\sqrt{2}}\right]\left[\frac{3\sqrt{2} -2}{2\sqrt{2}}\right]$

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

$=\frac{6-6\sqrt{2} +6\sqrt{2} -4}{8}$

$=\frac{2}{8}$

$=\frac{1}{4}$

Hence, $\quad\left(\cos 0^{\circ}+\sin 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right)=\frac{1}{4}$.