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

Question:

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

(a) $\frac{7}{4}$

(b) $\frac{5}{6}$

(c) $\frac{3}{5}$

(d) $\frac{5}{8}$

Solution:

As we know that,

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

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

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

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

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

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

By substituting these values, we get

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

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

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

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

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

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

$=\frac{7}{4}$

Hence, the correct option is (a).

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