Prove that the following equation is an identity.
esou
sin 4t/8= 1/2(cos^3tsint-sin^3tcost)

I suspect you mean

1/8 sin(4t ) = 1/2 (cos³(t ) sin(t ) - sin³(t ) cos(t ))

On the right side, pull out a factor of cos(t ) sin(t ):

1/2 (cos³(t ) sin(t ) - sin³(t ) cos(t )) = 1/2 cos(t ) sin(t ) (cos²(t ) - sin²(t ))

Recall the double angle identities for sin and cos :

sin(2t ) = 2 sin(t ) cos(t )

cos(2t ) = cos²(t ) - sin²(t )

Then

… = 1/4 (2 cos(t ) sin(t )) (cos²(t ) - sin²(t ))

… = 1/4 sin(2t ) cos(2t )

… = 1/8 (2 sin(2t ) cos(2t ))

… = 1/8 sin(4t )

understand that i don't know the answers

will someone answer his question