1 + cos ,{56^o} + cos ,{58^o} - cos {66^o} = | vedclass

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

Question

(c) $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $

$ = 2{\cos ^2}28^\circ + 2\sin 62^\circ .\sin 4^\circ $

$ = 2{\cos ^2}28^\circ + 2\cos

Vedclass · Accepted Answer

$4\,\cos {28^o}\,\cos \,{29^o}\,\sin {33^o}$

Vedclass · Answer

(c) $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $

$ = 2{\cos ^2}28^\circ + 2\sin 62^\circ .\sin 4^\circ $

$ = 2{\cos ^2}28^\circ + 2\cos 28^\circ .\sin 4^\circ $

$ = 2\cos 28^\circ (\cos 28^\circ + \cos 86^\circ )$

$ = 2\cos 28^\circ .2\cos 57^\circ \cos 29^\circ $

$ = 4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.

Aliter : Apply the conditional identity

$\cos A + \cos B - \cos C = - 1 + 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$                                                              $[\, \because 56^\circ  + 58^\circ  + 66^\circ  = 180^\circ ]$

We get the value of required expression equal to $4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.

$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $

Similar Questions

જો $A + B + C = \frac{\pi }{2}$ થાય તો $tanA\,\, tanB + tanB\,\, tanC + tanC\,\, tanA$ =

સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$

જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$

$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $

જો $x + y + z = {180^o},$ તો $\cos 2x + \cos 2y - \cos 2z = . . . .$