How do you verify the identity: cos^2x-sin^2x=1-2sin^2xcos2xsin2x=12sin2x?

4 Answers
Apr 22, 2015

From the basic definitions of sinsin and coscos and the Pythagorean Theorem
cos^2(x)+sin^2(x) = 1cos2(x)+sin2(x)=1

or
cos^2(x) = 1 -sin^2(x)cos2(x)=1sin2(x)

So
cos^2(x) - sin^2(x)cos2(x)sin2(x)

= (1-sin^2(x))-sin^2(x)=(1sin2(x))sin2(x)

= 1-2sin^2(x)=12sin2(x)

Apr 22, 2015

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

Here's one:

(use sin^2x+cos^2x = 1sin2x+cos2x=1

1-2sin^2x = (sin^2x + cos^2x) - 2sin^2x12sin2x=(sin2x+cos2x)2sin2x

color(white)"ssssssssssss"ssssssssssss =cos^2x+sin^2x-sin^2x-sin^2x=cos2x+sin2xsin2xsin2x

color(white)"ssssssssssss"ssssssssssss =cos^2x-sin^2x=cos2xsin2x

Apr 22, 2015

cos^2(x)-sin^2(x)cos2(x)sin2(x)

=(cos^2(x)+sin^2(x)) - (sin^2(x)+sin^2(x)) = 1 - 2sin^2(x)=(cos2(x)+sin2(x))(sin2(x)+sin2(x))=12sin2(x)

Jul 22, 2018

See below:

Explanation:

Recall the Pythagorean Identity

sin^2x+cos^2x=1sin2x+cos2x=1

Which can be manipulated into this form:

color(blue)(cos^2x=1-sin^2x)cos2x=1sin2x

In our equation, we can replace cos^2xcos2x with this to get

color(blue)(1-sin^2x)-sin^2x1sin2xsin2x, which simplifies to

1-2sin^2x12sin2x. 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!