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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 = $

Similar Questions

If $A$ and $B$ are values such that $(A+B)$ and $(A-B)$ are not odd multiples of $\frac{\pi}{2}$ and $2 \tan (A+B)=3 \tan (A-B)$,then $\sin 2A$ is equal to:

$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}=$

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

The value of $\sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$ is equal to

If $x = \sec \phi - \tan \phi$ and $y = \csc \phi + \cot \phi$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module