Question

Sin^8x - cos^8x = (sin^2 - cos^2x)(1-2sin^2xcos^2x)

Answer 1

Here use the formula, $a^2-b^2=(a+b)(a-b), a^4+b^4=(a^2+b^2)-2a^2b^2$

LHS is $\sin^8 x- \cos ^8 x=(\sin^4 x- \cos ^4 x)(\sin^4 x+ \cos ^4 x)\\ \sin^8 x- \cos ^8 x=(\sin^2 x+ \cos ^2 x)(\sin^2 x- \cos ^2 x)[(\sin^2 x+ \cos ^2 x)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(1)(\sin^2 x- \cos ^2 x)[(1)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(\sin^2 x- \cos ^2 x)(1-2\sin^2 x \cos ^2 x)\\ =RHS$

Answer 2

Answer:

Here use the formula, a2−b2=(a+b)(a−b),a4+b4=(a2+b2)−2a2b2a^2-b^2=(a+b)(a-b), a^4+b^4=(a^2+b^2)-2a^2b^2a

2

−b

2

=(a+b)(a−b),a

4

+b

4

=(a

2

+b

2

)−2a

2

b

2

LHS is sin⁡8x−cos⁡8x=(sin⁡4x−cos⁡4x)(sin⁡4x+cos⁡4x)sin⁡8x−cos⁡8x=(sin⁡2x+cos⁡2x)(sin⁡2x−cos⁡2x)[(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x]sin⁡8x−cos⁡8x=(1)(sin⁡2x−cos⁡2x)[(1)2−2sin⁡2xcos⁡2x]sin⁡8x−cos⁡8x=(sin⁡2x−cos⁡2x)(1−2sin⁡2xcos⁡2x)=RHS\begin{lgathered}\sin^8 x- \cos ^8 x=(\sin^4 x- \cos ^4 x)(\sin^4 x+ \cos ^4 x)\\ \sin^8 x- \cos ^8 x=(\sin^2 x+ \cos ^2 x)(\sin^2 x- \cos ^2 x)[(\sin^2 x+ \cos ^2 x)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(1)(\sin^2 x- \cos ^2 x)[(1)^2-2\sin^2 x \cos ^2 x]\\ \sin^8 x- \cos ^8 x=(\sin^2 x- \cos ^2 x)(1-2\sin^2 x \cos ^2 x)\\ =RHS\end{lgathered}

sin

8

x−cos

8

x=(sin

4

x−cos

4

x)(sin

4

x+cos

4

x)

sin

8

x−cos

8

x=(sin

2

x+cos

2

x)(sin

2

x−cos

2

x)[(sin

2

x+cos

2

x)

2

−2sin

2

xcos

2

x]

sin

8

x−cos

8

x=(1)(sin

2

x−cos

2

x)[(1)

2

−2sin

2

xcos

2

x]

sin

8

x−cos

8

x=(sin

2

x−cos

2

x)(1−2sin

2

xcos

2

x)

=RHS