Test Index

NCERT Class XI Mathematics - Trigonometric Functions - Solutions

© examsnet.com
Question : 41 of 61
Marks: +1, -0
cos 4x = 1 – 8 sin2sin^2 x cos2cos^2 x
Solution:  
We have, L.H.S.= cos 4x = cos 2(2x)
= 1 – 2 sin2sin^2 2x [Since cos2θ = 1 – 2 sin2θ]
= 1 – 2 (2sinxcosx)2(2 sinx cosx)^2 [Since sin 2θ = 2 sinθ cosθ]
= 1 – 2 [4 sin2sin^2 x cos2cos^2 x] = 1 – 8 sin2sin^2 x cos2cos^2 x = R.H.S.
© examsnet.com
Go to Question:
Prev Question
Next Question