Question

If `8x=π` then `sin^6(x) + cos^6(x) =`?

Answer 1

`sin^6x + cos^6x`

`=(sin^2x)^3 +( cos^2x)^3`

`=(sin^2x +cos^2x )^3-3xxsin^2xcos^2x(sin^2x+cos^2x)`

`=(1)^3-3/4xx(2sinxcosx)^2xx1`

`=1-3/4xx(sin2x)^2xx1`

`=1-3/4xx(sin(pi/4))^2`

`=1-3/4xx(1/sqrt2)^2`

`=1-3/4xx1/2=1-3/8=5/8`