Review question

# Can we write $\cos \theta - 2 \cos 3 \theta + \cos 5\theta$ using $\sin$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6120

## Solution

1. Prove the identity $\cos \theta - 2 \cos 3 \theta + \cos 5\theta = 2\sin \theta (\sin 2 \theta - \sin 4 \theta).$

We can always write the expressions $\cos A + \cos B, \cos A - \cos B, \sin A + \sin B$, and $\sin A - \sin B$ as the product of two $\sin / \cos$ terms.

This can be immensely useful to us in lots of situations. Here we are interested in $\cos A - \cos B$.

We begin with $\cos(X + Y) = \cos X \cos Y - \sin X \sin Y \quad \text{and} \quad \cos(X - Y) = \cos X \cos Y + \sin X \sin Y.$ Adding these equations gives $\cos (X-Y) - \cos (X+Y) = 2\sin X \sin Y.$ Now setting $A=X-Y, B=X+Y$ we find $\cos A - \cos B = 2 \sin \frac{A+B}2 \sin \frac{B-A}2.$

Applying this to both $\cos\theta - \cos 3\theta$ and $\cos 5 \theta - \cos 3\theta$, we see the left-hand side (LHS) is equal to $2 \sin 2\theta \sin \theta + 2 \sin 4\theta \sin (-2\theta) = 2 \sin \theta(\sin 2\theta - \sin 4 \theta),$ as required.

Alternatively, we can show by expanding that if $c = \cos\theta, s = \sin\theta$, then $\cos(3\theta) = 4c^3-3c, \quad \cos(5\theta) = 16c^5-20c^3+5c, \quad \sin(2\theta) = 2sc, \quad \sin(4\theta) = 8sc^3-4sc.$

Now the LHS is $c - 2(4c^3-3c)+ 16c^5-20c^3+5c=16c^5-28c^3+12c,$ while the RHS is $2s(2sc-(8sc^3-4sc)) = 2(1-c^2)(-8c^3+6c) = 16c^5-28c^3+12c,$ which are the same, and so we are done.

1. Solve the equations

1. $\cos 2x = \sin x$,

2. $3 \sec^2 x = \tan x + 5$

giving in each case all solutions between $0^\circ$ and $360^\circ$.

1. For the first equation, we use one of the double angle formulae for $\cos$, that’s $\cos 2x = 1 - 2\sin^2 x$.

Now everything’s written in terms of $\sin x$, which makes the equation much simpler.

Writing $u=\sin x$, we have to solve the equation $\cos 2x = \sin x,$ which gives $1 - 2u^2 = u, \quad \text{ or} \quad 2u^2+u-1=0, \quad \text{ or} \quad (2u-1)(u+1)=0,$
with solutions $\sin x = u = \frac{1}{2}, -1.$

We need all solutions between $0^\circ$ and $360^\circ$. These are $x = 30^\circ, 150^\circ, 270^\circ.$

It’s helpful to sketch a graph, to check we’ve got all the solutions.

1. For the second, recall the identity $\tan^2 x + 1 = \sec^2 x$ (given by dividing $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$).

Define $v = \tan x$, so we need to solve

$3 \sec^2 x = \tan x + 5, \quad \text{ or} \quad 3(1+v^2)-v-5=0, \quad \text{ or} \quad 3v^2 -v -2=0.$ This is $(3v+2)(v-1)=0$, so $v = -\frac {2}{3}, 1$, which (since $\tan x$ has period $\pi$) gives the four solutions $x = 45^\circ, 225^\circ, 146.3^\circ, 326.3^\circ.$