Math, asked by PragyaTbia, 11 months ago

निम्नलिखित को सिद्ध कीजिए: \cos^2{2x} - \cos^2{6x} = \sin 4x\,\sin 8x

Answers

Answered by jagadeep72
0

Mark this as brainliest answer

Attachments:
Answered by kaushalinspire
0

Answer:

Step-by-step explanation:

cos^{2} 2x - cos^{2} 6x  =  sin4x sin8x

L.H.S.    =   [tex]cos^{2} 2x - cos^{2} 6x

            =   1 - sin^{2}2x - ( 1 - sin^{2}6x )

             =  - sin^{2}2x + sin^{2}6x

             =   sin^{2}6x - sin^{2}2x

माना कि   6x = A   तथा   2x = B

अतः  सूत्र   sin^{2}A - sin^{2}B  =  sin(A+B) sin(A-B)  के अनुसार

 L.H.S.  =  sin^{2}6x - sin^{2}2x

             =  sin(6x+2x) . sin(6x-2x)

             =  sin8x . sin4x

             =  sin4x . sin8x  =  R.H.S.

Similar questions