Is this an identity, if so how can it be broken down? #[1 + sin(y)] [ 1 + sin(-y)] = cos^2(y)#

1 Answer
Feb 18, 2018

Using these known identities:

#color(white){color(black)( (sin(-theta)=-sintheta, qquad(1)), (sin^2theta+cos^2theta=1, qquad(2.1)), (cos^2theta=1-sin^2theta, qquad(2.2) ) :}#

(Identity #(2.2)# is achieved by subtracting #sin^2theta# from both sides of identity #(2.1)#.)

We can prove the identity (I will be manipulating the left side of the equation until it equals the right side):

#color(white){color(black)( ((1+siny)(1+sin(-y)), qquad"Left hand side"), ((1+siny)(1-siny), qquad"Use identity "(1)), (1-siny+siny-sin^2y, qquad"Use FOIL method"), (1color(red)cancel(color(black)(-siny+siny))-sin^2y, qquad"Cancel like terms"), (1-sin^2y, qquad"Rewrite"), (cos^2y, qquad"Use identity " (2.2)):}#

That's it! We just proved the identity because the left side equals the right side.