### Trigonometry: Compound Angles

Many ways problem

## Things you might have noticed

Take a look at these identities.

$\cos^2 \frac{\theta}{2} \equiv \frac{1}{2}(1+\cos \theta) \quad \quad \quad \sin^2 \frac{\theta}{2} \equiv \frac{1}{2}(1-\cos \theta)$

How could you use these identities to help you sketch graphs of $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$?

These ideas may help you to sketch the graphs of $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$

• As the left-hand side is a square in each case, we know the right hand sides must be zero or positive. This is fine, because $-1\leq \cos \theta \leq 1$, so $1+\cos \theta$ will be at least $0$ and at most $2$. Similarly, $1-\cos \theta$ must be between $0$ and $2$.

• The right-hand sides of the identities are the same except for a minus sign, so we’d expect the behaviour of the two functions to be very similar, but the maximum and minimum values will occur at different values of $\theta$. For example, $\frac{1}{2}(1+\cos \theta)$ will be $1$ when $\theta = \dotsc ,-4\pi, -2\pi, 0, 2\pi,4\pi, \dotsc$ (i.e. at even multiples of $\pi$) whereas $\frac{1}{2}(1-\cos \theta)$ will be $1$ when $\theta = \dotsc,-5\pi, -3\pi, -\pi, \pi, 3\pi,5\pi, \dotsc$ (i.e. at odd multiples of $\pi$).

• The right-hand side of each identity only involves $\cos \theta$ and constants, so the graphs of $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$ can be thought of as transformations of a $y=\cos \theta$ graph. Which transformations are involved in each case? What is the frequency of $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$?

• Again, from the right-hand side, the identities tell us that $\cos^2 \frac{\theta}{2}$ is the mean of $1$ and $\cos \theta$ and $\sin^2 \frac{\theta}{2}$ is the mean of $1$ and $-\cos \theta$. What does this tell you about the graph of $y= \cos^2 \frac{\theta}{2}$ in relation to the graphs of $y=1$ and $y=\cos \theta$?