The value of cos 20^{circ} + 2 sin^2 55^{circ | vedclass

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(B) Given expression: $\cos 20^{\circ} + 2 \sin^2 55^{\circ} - \sqrt{2} \sin 65^{\circ}$ Using the identity $2 \sin^2 	heta = 1 - \cos 2	heta$,we ha

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(B) Given expression: $\cos 20^{\circ} + 2 \sin^2 55^{\circ} - \sqrt{2} \sin 65^{\circ}$ Using the identity $2 \sin^2 	heta = 1 - \cos 2	heta$,we have $2 \sin^2 55^{\circ} = 1 - \cos 110^{\circ}$. Substituting this into the expression: $\cos 20^{\circ} + 1 - \cos 110^{\circ} - \sqrt{2} \sin 65^{\circ}$ Since $\cos 110^{\circ} = \cos(90^{\circ} + 20^{\circ}) = -\sin 20^{\circ}$,the expression becomes: $\cos 20^{\circ} + 1 + \sin 20^{\circ} - \sqrt{2} \sin 65^{\circ}$ Alternatively,using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$: $\cos 20^{\circ} - \cos 110^{\circ} = 2 \sin \frac{110^{\circ}+20^{\circ}}{2} \sin \frac{110^{\circ}-20^{\circ}}{2} = 2 \sin 65^{\circ} \sin 45^{\circ}$ $= 2 \sin 65^{\circ} 	imes \frac{1}{\sqrt{2}} = \sqrt{2} \sin 65^{\circ}$ Substituting back: $\sqrt{2} \sin 65^{\circ} - \sqrt{2} \sin 65^{\circ} + 1 = 1$

The value of $\cos 20^{\circ} + 2 \sin^2 55^{\circ} - \sqrt{2} \sin 65^{\circ}$ is

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