How do you solve cos^2 theta=2cos^2 (theta/2)cos2θ=2cos2(θ2) in the domain 0<=theta<=3600θ360?

2 Answers
Jun 26, 2018

The solutuions are S={128.2^@, 231.8^@}S={128.2,231.8}

Explanation:

We need

costheta=2cos^2(theta/2)-1cosθ=2cos2(θ2)1

-1<=costheta<=11cosθ1

Then, the equation is

cos^2theta=2cos^2(theta/2)cos2θ=2cos2(θ2)

cos^2theta=costheta+1cos2θ=cosθ+1

cos^2theta-costheta-1=0cos2θcosθ1=0

This is a quadratic equation in cos^2thetacos2θ

The solutions are

costheta=(1+-sqrt((-1)^2-4(1)(-1)))/(2)cosθ=1±(1)24(1)(1)2

=(1+-sqrt5)/2=1±52

{(costheta=1.62),(costheta=-0.618):}

=>, {(theta=O/),(theta=128.2^@ ; 231.8^@):}

\theta=128.18^\circ, 231.82^\circ

Explanation:

Given that

\cos^2\theta=2\cos^2\frac{\theta}{2}

(2\cos^2\frac{\theta}{2}-1)^2=2\cos^2\frac{\theta}{2}

4\cos^4\frac{\theta}{2}+1-4\cos^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}

4\cos^4\frac{\theta}{2}-6\cos^2\frac{\theta}{2}+1=0

Solving above quadratic equation in terms of \cos^2\frac{\theta}{2} using quadratic formula as follows

\cos^2\frac{\theta}{2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(4)(1)}}{2(4)}

\cos^2\frac{\theta}{2}=\frac{3\pm\sqrt{5}}{4}
But , 0\le\cos^2\frac{\theta}{2}\le1

\therefore \cos^2\frac{\theta}{2}=\frac{3-\sqrt{5}}{4}

\cos^2\frac{\theta}{2}=(\frac{\sqrt5-1}{2\sqrt2})^2

Let \cos^2\frac{\theta}{2}=\cos^2\alpha

\frac{\theta}{2}=n\pi\pm\alpha

\theta=2(n\pi\pm\alpha)

Where, \cos\alpha=\frac{\sqrt5-1}{2\sqrt2}=64.09^\circ &
n is any integer i.e. n=0, \pm1, \pm2, \pm3, \ldots

But since, 0\le\theta\le360^\circ hence, setting n=0\ &\ 1 & taking positive & negative signs respectively, we get two values of \theta as follows

\theta=2(0+64.09^\circ), 2(180^\circ-64.09^\circ)
=128.18^\circ, 231.82^\circ