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