# Evaluate each of the following

Question:

Evaluate each of the following

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

Solution:

We have,

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

Now,

$\sin 90^{\circ}=\cos 0^{\circ}=1, \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}$

So by substituting above values in equation (1)

We get,

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

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

Now, LCM of both the product terms in the above expression is $2 \sqrt{2}$

Therefore we get,

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

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

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

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

Now by rearranging terms in the numerator of above expression

We get,

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

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

$=\frac{(2 \sqrt{2}+\sqrt{2}+2) \times(2 \sqrt{2}+\sqrt{2}-2)}{(2 \sqrt{2}) \times(2 \sqrt{2})}$

Now, by applying formula $\left[(a+b)(a-b)=a^{2}-b^{2}\right]$ in the numerator of the above expression we get,

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

$=\frac{(2 \sqrt{2}+\sqrt{2})^{2}-2^{2}}{2 \times 2 \times \sqrt{2} \times \sqrt{2}}$

$=\frac{(2 \sqrt{2}+\sqrt{2})^{2}-2^{2}}{4 \times 2}$....(2)

Now, we know that $(a+b)^{2}=a^{2}+2 a b+b^{2}$

Therefore,

$(2 \sqrt{2}+\sqrt{2})^{2}=(2 \sqrt{2})^{2}+2(2 \sqrt{2})(\sqrt{2})+(\sqrt{2})^{2}$

Now, by substituting the above value of $(2 \sqrt{2}+\sqrt{2})^{2}$ in equation (2)

We get,

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

$=\frac{\left[(2 \sqrt{2})^{2}+2(2 \sqrt{2})(\sqrt{2})+(\sqrt{2})^{2}\right]-2^{2}}{4 \times 2}$

$=\frac{[8+8+2]-4}{8}$

$=\frac{18-4}{8}$

$=\frac{14}{8}$

Now $\frac{14}{8}$ gets reduced to $\frac{7}{4}$

Therefore,

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

$=\frac{7}{4}$

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