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

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

1 Answer
Feb 28, 2017

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!