Question

prove that cos^8-sin^8 = (cos^2-sin^2) (1-2sin^2cos^2)​

Answer 1

(sin

8

θ−cos

8

θ)=(sin

4

θ)

2

−(cos

4

θ)

2

=(sin

4

θ−cos

4

θ)(sin

4

θ+cos

4

θ)

⇒LHS=(sin

2

θ−cos

2

θ)(sin

2

θ+cos

2

θ)(sin

4

θ+cos

4

θ)

⇒LHS=(sin

2

θ−cos

2

θ)[(sin

2

θ)

2

+(cos

2

θ)

2

+2sin

2

θcos

2

θ−2sin

2

θcos

2

θ]

⇒LHS=(sin

2

θ−cos

2

θ)[(sin

2

θ+cos

2

θ)

2

−2sin

2

θcos

2

θ]

⇒LHS=(sin

2

θ−cos

2

θ)(1−2sin

2

θcos

2

θ)=RHS