1 + cos 56^circ + cos 58^circ - cos 66^circ = | vedclass

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Question

(C) Given expression: $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $Using the identity $1 + \cos 2	heta = 2 \cos^2 	heta$,we have $1 + \cos 56

Vedclass · Accepted Answer

$4 \cos 28^\circ \cos 29^\circ \sin 33^\circ $

Vedclass · Answer

(C) Given expression: $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $Using the identity $1 + \cos 2	heta = 2 \cos^2 	heta$,we have $1 + \cos 56^\circ = 2 \cos^2 28^\circ$.Now,the expression becomes $2 \cos^2 28^\circ + (\cos 58^\circ - \cos 66^\circ)$.Using the identity $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$,we get $\cos 58^\circ - \cos 66^\circ = 2 \sin \frac{58^\circ + 66^\circ}{2} \sin \frac{66^\circ - 58^\circ}{2} = 2 \sin 62^\circ \sin 4^\circ$.Since $\sin 62^\circ = \cos 28^\circ$,the expression is $2 \cos^2 28^\circ + 2 \cos 28^\circ \sin 4^\circ = 2 \cos 28^\circ (\cos 28^\circ + \sin 4^\circ)$.Since $\sin 4^\circ = \cos 86^\circ$,we have $2 \cos 28^\circ (\cos 28^\circ + \cos 86^\circ)$.Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get $2 \cos 28^\circ [2 \cos \frac{28^\circ + 86^\circ}{2} \cos \frac{86^\circ - 28^\circ}{2}] = 2 \cos 28^\circ [2 \cos 57^\circ \cos 29^\circ]$.Since $\cos 57^\circ = \sin 33^\circ$,the final result is $4 \cos 28^\circ \cos 29^\circ \sin 33^\circ$.

$1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ = $

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