what is 2cos^2(θ)−1
I assume you want to solve this:
2cos^2(θ)−1 = 0
Add 1 to both sides
2cos^2(θ) =1
Divide by 2 on both sides
cos^2(θ) = 1/2
Take the square root of both sides
cos(θ)= ± 1/√2
So θ = pi/4, 3pi/4, 5pi/4 and 7pi/4 in the interval [0, 2pi]
what is 2cos^2(θ)−1 ?
$$\cos{(2\theta)}=\cos^2{(\theta)} -\underbrace{ \sin^2{(\theta)}}_{=( 1- \cos^2{(\theta)} ) }\\ \cos{(2\theta)}=\cos^2{(\theta)} -1 +\cos^2{(\theta)} \\ \cos{(2\theta)} =2\cos^2{(\theta)} -1 \\\\ \boxed{ 2\cos^2{(\theta)} -1 = \cos{(2\theta)} }$$