# (cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

Question:

(cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

Solution:

As we know that,

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

$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

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

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

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

By substituting these values, we get

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

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

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

$=1+\frac{1}{4}+2(1)\left(\frac{1}{2}\right)-\frac{1}{2}$

$=1+\frac{1}{4}+1-\frac{1}{2}$

$=\frac{4+1+4-2}{4}$

$=\frac{7}{4}$

Hence, $\left(\cos 0^{\circ}+\cos 45^{\circ}+\sin 30^{\circ}\right)\left(\sin 90^{\circ}-\cos 45^{\circ}+\cos 60^{\circ}\right)=\frac{7}{4}$.