Question

prove sin 8 theta minus Cos 8 theta equal to sin square theta minus cos square theta 1 - 2 sin square theta cos square theta

Answer 1

hello there !!
kindly post more such questions cuz i love solving 'em

note i will be using A instead of theta
we have,
LHS= sin^8-cos^8
=> (sin⁴A)²-(cos⁴A)²

=>(sin⁴A-cos⁴)(sin⁴A+cos⁴A) [Using algebraic identity (a+b)(a-b)=a²-b²]

=>(sin²A-cos²A)(sin²A+cos²A)(sin⁴A+cos⁴A) [applied same identity]

=> (sin²A-cos²A)[(sin²A)²+(cos²A)] [since sin²A+cos²A=1]

=>(sin²A-cos²A)[(sin²A)²+(cos²A) +2sin²Acos²A-2sin²Acos²A]
[adding and subtracting 2sin²Acos²A]

=> (sin²A-cos²A)[(sin²A+cos²A)²-2sin²Acos²A]

=>(sin²A-cos²A)[(1)²-2sin²Acos²A]

=> (sin²A-cos²A)(1-2sin²Acos²A)
LHS=RHS
HOPE THIS HELPED YOU ;P

Answer 2

Answer:

sin⁸θ-cos⁸θ

=(sin⁴θ)²-(cos⁴θ)²

=(sin⁴θ+cos⁴θ)(sin⁴θ-cos⁴θ)

={(sin²θ)²+(cos²θ)²}{(sin²θ)²-(cos²θ)²}

={(sin²θ+cos²θ)²-2sin²θcos²θ}{(sin²θ+cos²θ)(sin²θ-cos²θ)}

={(1)²-2sin²θcos²θ}{(1)(sin²θ-cos²θ)} [∵, sin²θ+cos²θ=1]

=(sin²θ-cos²θ)(1-2sin²θcos²θ) (Proved)

If helped please mark as brainliest

Thanks