Question

Enter the expression 2cos2(θ)−1 , where θ is the lowercase Greek letter theta.

Answer 1

`2cos^2(\theta) - 1 = cos(2\theta)`

Step-by-step explanation:

Given

`2cos^2(\theta) - 1`

Required

Simplify

In trigonometry:

`sin^2(\theta) + cos^2(\theta) = 1`

So; the given expression becomes

`2cos^2(\theta) - (sin^2(\theta) + cos^2(\theta))`

Open Bracket

`2cos^2(\theta) - sin^2(\theta) - cos^2(\theta)`

Collect Like Terms

`2cos^2(\theta) - cos^2(\theta)- sin^2(\theta)`

`cos^2(\theta)- sin^2(\theta)`

In trigonometry:

`cos(\theta + \theta) = cos^2(\theta)- sin^2(\theta)`

This implies that:

`cos^2(\theta)- sin^2(\theta) = cos(\theta + \theta)`

=

`cos(\theta + \theta)`

`cos(2\theta)`

Hence:

`2cos^2(\theta) - 1 = cos(2\theta)`