The value of cos ^2 10^{circ}-cos 10^{circ} c | vedclass

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(C) Let $E = \cos ^2 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^2 50^{\circ}$.Using the identity $\cos ^2 	heta = \frac{1+\cos 2	heta}{2}$,we

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) Let $E = \cos ^2 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^2 50^{\circ}$.Using the identity $\cos ^2 	heta = \frac{1+\cos 2	heta}{2}$,we have:$E = \frac{1+\cos 20^{\circ}}{2} - \cos 10^{\circ} \cos 50^{\circ} + \frac{1+\cos 100^{\circ}}{2}$$E = \frac{1}{2} + \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} [2 \cos 50^{\circ} \cos 10^{\circ}] + \frac{1}{2} + \frac{1}{2} \cos 100^{\circ}$Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$E = 1 + \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} [\cos 60^{\circ} + \cos 40^{\circ}] + \frac{1}{2} \cos 100^{\circ}$$E = 1 + \frac{1}{2} \cos 20^{\circ} - \frac{1}{4} - \frac{1}{2} \cos 40^{\circ} + \frac{1}{2} \cos 100^{\circ}$$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} + \frac{1}{2} [\cos 100^{\circ} - \cos 40^{\circ}]$Using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} + \frac{1}{2} [-2 \sin 70^{\circ} \sin 30^{\circ}]$Since $\sin 70^{\circ} = \cos 20^{\circ}$ and $\sin 30^{\circ} = \frac{1}{2}$:$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} - \cos 20^{\circ} \cdot \frac{1}{2} = \frac{3}{4}$.

The value of $\cos ^2 10^{\circ}-\cos 10^{\circ} \cdot \cos 50^{\circ}+\cos ^2 50^{\circ}$ is:

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