How do you prove #(1+cosx)(1-cosx)=sin^2x#?

2 Answers
Mar 14, 2018

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

Explanation:

the identity known is #sin^2x + cos^2x = 1#.

this can be rearranged to give #1 - cos^2x = sin^2x#.

using the 'difference of two squares' identity,

where #(a+b)(a-b) = a^2-b^2#,

#(1+cosx)(1-cosx) = 1^2 - cos^2x#

#1^2 = 1#

#(1+cosx)(1-cosx) = 1 - cos^2x#

since #1 - cos^2x = sin^2x#, #(1+cosx)(1-cosx) = sin^2x#.

Mar 14, 2018

FOIL the LHS and use the trig identity that #sin^2x+cos^2x=1#

Explanation:

First, FOIL the left hand side (LHS):

#(1+cosx)(1-cosx)=1cancel(-cosx+cosx)+cos^2x#

#=1-cos^2x#

Now, re-write the equation:

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

There is a trig identity that states:

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

Substitute the 1 in our proof:

#sin^2xcancel(+cos^2x-cos^2x)=sin^2x#

#sin^2x=sin^2x#