cos ^2 10^{circ}+cos ^2 50^{circ}-sin 40^{cir | vedclass

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Question

(D) We have the expression: $\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ Using the identity $2 \cos ^2 A = 1 + \cos 2A$ and

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(D) We have the expression: $\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ Using the identity $2 \cos ^2 A = 1 + \cos 2A$ and $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $= \frac{1}{2} [ (1 + \cos 20^{\circ}) + (1 + \cos 100^{\circ}) - (\cos(40^{\circ}-80^{\circ}) - \cos(40^{\circ}+80^{\circ})) ]$ $= \frac{1}{2} [ 2 + \cos 20^{\circ} + \cos 100^{\circ} - \cos(-40^{\circ}) + \cos 120^{\circ} ]$ Since $\cos(-	heta) = \cos 	heta$ and $\cos 120^{\circ} = -\frac{1}{2}$: $= \frac{1}{2} [ 2 + \cos 20^{\circ} + \cos 100^{\circ} - \cos 40^{\circ} - \frac{1}{2} ]$ $= \frac{1}{2} [ \frac{3}{2} + (\cos 100^{\circ} + \cos 20^{\circ}) - \cos 40^{\circ} ]$ Using $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $= \frac{1}{2} [ \frac{3}{2} + 2 \cos 60^{\circ} \cos 40^{\circ} - \cos 40^{\circ} ]$ Since $\cos 60^{\circ} = \frac{1}{2}$: $= \frac{1}{2} [ \frac{3}{2} + 2(\frac{1}{2}) \cos 40^{\circ} - \cos 40^{\circ} ]$ $= \frac{1}{2} [ \frac{3}{2} + \cos 40^{\circ} - \cos 40^{\circ} ] = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$

$\cos ^2 10^{\circ}+\cos ^2 50^{\circ}-\sin 40^{\circ} \sin 80^{\circ}$ is equal to

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