Question

Prove that
sin8theeta-cos8theeta = (sin2theeta-cos2theeta)(1-2sin2theeta.cos2theeta)​

Answer 1

Step-by-step explanation:

We have ,

(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