Question

Derivative of the sine and cosine functions. First Derivative. How can I simplify this expression ?

y=`sin^2`( `pi`-x)

Answer 1

The first derivative equals `sin2x`.

Explanation:

This is equivalent to saying

`y = (sin(pi - x))(sin(pi - x))`

We can expand `sin(pi - x)` using `sin(A - B) = sinAcosB - sinBcosA`

`y = (sinpicosx - sinxcospi)(sinpicosx - sinxcospi)`

`y = (0(cosx) - sinx(-1))((0)cosx - (-1)sinx)`

`y= (sinx)(sinx)`

`y = (sinx)^2`

Now use the chain rule to differentiate. We let `y = u^2` and `u = sinx`. Then `(dy)/(du) = 2u` and `(du)/dx = cosx`.

`y' = 2u * cosx`

`y' = 2sinxcosx`

`y' = sin2x`

Hopefully this helps!