How do you verify the identity: #cos^2x-sin^2x=1-2sin^2x#?

4 Answers
Apr 22, 2015

From the basic definitions of #sin# and #cos# and the Pythagorean Theorem
#cos^2(x)+sin^2(x) = 1#

or
#cos^2(x) = 1 -sin^2(x)#

So
#cos^2(x) - sin^2(x)#

#= (1-sin^2(x))-sin^2(x)#

#= 1-2sin^2(x)#

Apr 22, 2015

There are many acceptable (correct) ways to do this.

Here's one:

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

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

#color(white)"ssssssssssss"# #=cos^2x+sin^2x-sin^2x-sin^2x#

#color(white)"ssssssssssss"# #=cos^2x-sin^2x#

Apr 22, 2015

#cos^2(x)-sin^2(x)#

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

Jul 22, 2018

Answer:

See below:

Explanation:

Recall the Pythagorean Identity

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

Which can be manipulated into this form:

#color(blue)(cos^2x=1-sin^2x)#

In our equation, we can replace #cos^2x# with this to get

#color(blue)(1-sin^2x)-sin^2x#, which simplifies to

#1-2sin^2x#. We have just verified the identity

#bar( ul(|color(white)(2/2)cos^2x-sin^2x=1-2sin^2x color(white)(2/2)|))#

With the use of the Pythagorean Identity.

Hope this helps!