Question

Prove that `(sin4t)/4 -= cos^3tsint-sin^3tcost`?

Answer 1

We seek to show that:

`(sin4t)/4 -= cos^3tsint-sin^3tcost`

Consider the LHS:

`LHS = (sin4t)/4`

Using the sine and cosine double angle formulas:

`sin2A -= 2sinAcosA`
`cos2A -= cos^2A-sin^2A`

we have:

`LHS = (sin(2(2t)))/4`

`\ \ \ \ \ \ \ \ = (2sin2tcos2t)/4`

`\ \ \ \ \ \ \ \ = (2(2sintcost)(cos^2t-sin^2t))/4`

`\ \ \ \ \ \ \ \ = sintcost(cos^2t-sin^2t)`

`\ \ \ \ \ \ \ \ = sintcostcos^2t-sintcostsin^2t`

`\ \ \ \ \ \ \ \ = sintcos^3t-costsin^3t`

`\ \ \ \ \ \ \ \ = RHS \ \ \ \` QED