Question

64 ( \cos ^ { 8 } \theta + \sin ^ { 8 } \theta ) = \cos \theta + 28 \cos ^ { 4 } \theta + 35

Answer 1

Answer:

We have ,

We have , (sin 8 θ−cos 8 θ)=(sin 4 θ) 2 −(cos 4 θ) 2

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]

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

HOPE IT'S CLEAR

PLEASE MARK ME AS BRAINLIEST