How do you verify #(1+sinx+cosx)^2 = 2(1+sinx)(1+cosx)#?

1 Answer
Jun 15, 2015

You will be squaring a quantity that contains both #sinx# and #cosx#, so you will need the identity #sin^2x+cos^2x = 1#. I would start by multiplying both of these out.

#(1+sinx+cosx)^2 = 1 + sinx + cosx + sinx + sin^2x + sinxcosx + cosx + sinxcosx + cos^2x#

#= 1 + 2sinx + 2cosx + sin^2x + cos^2x + 2sinxcosx#

while

#2(1+sinx)(1+cosx) = 2(1+cosx+sinx+sinxcosx) = 2+2cosx+2sinx+2sinxcosx#

Compare:
#1 + sin^2x + cos^2x + cancel(2sinx + 2cosx + 2sinxcosx) = 2+cancel(2cosx+2sinx+2sinxcosx)#

#1 + sin^2x + cos^2x = 2#

#1 + 1 = 2#

#2 = 2#

They're equal.