Question

Cos4x = 1-8sin square x cos2x

Answer 1

Answer:

Given cos4x=1−8sin

2

xcos

2

x

Taking LHS

cos4x

We know

cos2x=2cos

2

x−1

Replacing x by 2x

cos2(2x)=2cos

2

(2x)−1

cos4x=2cos

2

2x−1

=2(cos2x)

2

−1

Using cos2x=2cos

2

x−1

=2(2cos

2

x−1)

2

−1

Using (a−b)

2

=a

2

+b

2

−2ab

=2[(2cosx)

2

+(1)

2

−2(2cos

2

x)×1]−1

=2(4cos

4

x+1−4cos

2

x)−1

=2×4cos

4

x+2−2×4cos

2

x−1

=8cos

4

x+2−8cos

2

x−1

=8cos

4

x−8cos

2

x+2−1

=8cos

2

x(cos

2

x−1)+1

=8cos

2

x[−(1−cos

2

x)]+1

=−8cos

2

x[(1−cos

2

x)]+1

=−8cos

2

xsin

2

x+1 [∵sin

2

x=1−cos

2

x]

=1−8cos

2

xsin

2

x

=RHS