# Simple trigonometric problem

• John O' Meara

## Homework Statement

$$\cos (2 \theta)=- \cos ( \theta)$$

## The Attempt at a Solution

In general $$\theta = 2n \pi +/- \alpha \mbox{ where } \cos \theta = \cos \alpha$$
Because $$\cos(2 \theta) = - \cos( \theta) \mbox{ then, } 2 \theta = 2n \pi -/+ \theta$$
taking $$2 \theta = 2n \pi - \theta \mbox{ that implies } 3 \theta = 2n \pi \mbox{ therefore } \theta = \frac{2}{3}n \pi$$. The book says that theta must lie between 0 and pi/2. So I must have gone wrong. Can anyone put me right on this equation. Thanks for the help.

This is confusing. Are you trying to show that the left-hand side is equal to the right-hand side (which I don't think is true) or are you trying to solve some other problem?

I am trying to show that the left side equals the right, but I am not sure if I am correct. The equation itself results from solving another problem. Thanks.

cos(theta) = -cos(alpha) implies that either

theta = alpha +2n*pi

or

theta= (pi-alpha) +2n*pi.

ehild

$$-cos\theta=cos(\pi+\theta)$$
$$2\theta_{n,+}=\left(2n+1\right)\pi+\theta_{n,+}$$
$$2\theta_{n,-}=\left(2n+1\right)\pi-\theta_{n,-}$$
$$\theta_{n,+}=\left(2n+1\right)\pi$$
$$\theta_{n,-}=\frac{\left(2n+1\right)\pi}3$$
There are no solutions $$\theta_{n,+}$$ within the interval $$\left[0,\frac \pi 2\right]$$, but the solution $$\theta_{0,-}=\frac \pi 3$$ exists.

(Edit: Yeah, what they said.)

No, the equation I want solved is $$\cos (2 \theta)= - \cos( \theta)$$. I do not know how to do it. Thanks.

Last edited:
Thanks Gigasoft, and thank you all.

What you need is the identity $cos(2\theta)= cos^2(\theta)- sin^2(\theta)=$$cos^2(\theta)- (1- cos^2(\theta))$

$cos(2\theta)= 2cos^2(\theta)- 1$.

You get a quadratic equation in $cos(\theta)$.