Review question

# Can we solve $\cos\theta - \sin (2\theta) + \cos (3\theta) - \sin (4\theta) = 0$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8648

## Solution

Give the general solutions of the following equations

1. $2\sin 3\theta - 7 \cos 2\theta + \sin \theta + 1 = 0$,

We’ll try to write everything in terms of just $\sin \theta$ (the single $\sin\theta$ in the LHS points to this). Let $s = \sin \theta$, and $c = \cos\theta$.

We could use the double-angle formula $\cos 2\theta = 1 - 2s^2.$ We’d like to expand $\sin 3\theta$. Using the angle addition formula,

\begin{align*} \sin 3\theta &= c\sin 2\theta + s\cos 2\theta\\ &= 2sc^2 +s(1-2s^2)\\ &= 2s(1-s^2) + s(1-2s^2)\\ &= 3s - 4s^3. \end{align*} The original equation becomes \begin{align*} 2(3s - 4s^3) - 7(1-2s^2) + s +1 &= 0\\ \implies 8s^3 - 14s^2 - 7s + 6 &= 0. \end{align*}

When solving a cubic equation, it is often sensible to try simple numbers like $s= \pm 1, \pm 2$ to find the first factor using the Factor Theorem.

If we try small integers for $s$, we find that $8s^3 - 14s^2 - 7s + 6 = 0$ when $s = 2$, and so by the Factor Theorem, $s-2$ is a factor. Thus

$8s^3 - 14s^2 - 7s + 6 = (s-2)(8s^2+2s -3) =(s-2)(4s+3)(2s-1),$

which is zero when $s = 2, -\dfrac{3}{4}$ or $\dfrac{1}{2}$.

We can note that $\sin \theta = 2$ is impossible.

As the period of $\sin$ is $2\pi$, and $\sin(\theta) = \sin(\pi-\theta)$, the remaining solutions are \begin{align*} \theta = \quad & \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta = \quad & \pi - \sin^{-1}\left(\frac 1 2\right) + 2\pi n, \\ \theta =\quad & \sin^{-1}\left(- \frac 3 4\right) + 2\pi n, \\ \theta =\quad & \pi-\sin^{-1}\left(- \frac 3 4\right) + 2\pi n \end{align*} where $n$ is an integer. We can simplify to find \begin{align*} \theta =\quad & \dfrac{\pi}{6} +2n\pi, \\ \theta =\quad & \dfrac{5\pi}{6} +2n\pi, \\ \theta =\quad & -\sin^{-1}\left(\frac 3 4\right)+2n\pi, \\ \theta =\quad & \sin^{-1}\left(\frac 3 4\right)+(2n+1)\pi, \end{align*}

where in each case $n$ is any integer.

1. $\cos\theta - \sin 2\theta + \cos 3\theta - \sin 4\theta = 0$.

We might try an approach similar to that above, by writing everything in terms of $\cos \theta$ and $\sin \theta$. It’s probably quicker, however, to use the sum-to-product formulae, which help us to combine two $\sin$ or two $\cos$ functions.

$\sin A + \sin B = 2 \sin\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right)$ and $\cos A + \cos B = 2 \cos\left(\frac{A+B} 2\right) \cos\left(\frac{A-B} 2\right).$

We can derive these simply from \begin{align*} \sin(X\pm Y) &= \sin(X)\cos(Y) \pm \cos(X)\sin(Y),\\ \cos(X\pm Y) &= \cos(X)\cos(Y) \mp \sin(X)\sin(Y) \end{align*}
Applying these to this part of the question we obtain $\cos \theta + \cos 3\theta = 2 \cos 2\theta \cos \theta , \qquad \sin 2\theta + \sin 4\theta = 2 \sin 3\theta \cos \theta$ and so we need to solve \begin{align*} 2\cos 2\theta \cos \theta - 2\sin3\theta\cos\theta & = 0 \\ \Longleftrightarrow \cos \theta (\cos 2 \theta - \sin 3\theta) & = 0. \end{align*}

So our solutions are $\cos \theta = 0$, which gives $\theta = \dfrac{\pi}{2} + n\pi$, or are the solutions to $\cos 2\theta = \sin 3\theta$.

Now $\cos 2\theta = 1-2s^2$, while (as we showed above) $\sin 3\theta = 3s-4s^3$, so we need to solve \begin{align*} 1-2s^2&=3s-4s^3\\ \implies 4s^3-2s^2-3s+1&=0\\ \implies (s-1)(4s^2+2s -1)&=0\\ \implies \sin \theta = 1 \quad \text{or} \quad \sin \theta &= \dfrac{-1\pm\sqrt{5}}{4}. \end{align*} So our final solution is \begin{align*} \theta = \quad &\dfrac{\pi}{2} + n\pi\\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1+ \sqrt{5}}{4}\right) + 2n\pi, \\ \text{or} \quad \theta = \quad &\sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi,\\ \text{or} \quad \theta = \quad &\pi - \sin^{-1}\left(\dfrac{-1- \sqrt{5}}{4}\right) + 2n\pi, \end{align*}

where $n$ is an integer.

Alternatively, we could note that $\cos \alpha = \cos \beta \implies \alpha = \pm \beta + 2n\pi.$

We also have (by considering a right-angled triangle where one angle is $\gamma$) that $\sin \gamma = \cos\left(\dfrac{\pi}{2}-\gamma\right).$ This is in fact true for all $\gamma$.

Thus the equation $\cos 2\theta = \sin 3\theta$ becomes \begin{align*} \cos 2\theta &= \cos \left(\dfrac{\pi}{2}-3\theta \right)\\ \implies 2\theta &= \pm \left(\dfrac{\pi}{2}-3\theta\right) +2n\pi\\ \implies \theta &= \dfrac{\pi}{2} +2n\pi \quad \text{or} \quad \dfrac{\pi}{10} + \dfrac{2n\pi}{5}, \end{align*}

and so we have the same answers as before, but this time in an explicit form.

Notice we have inadvertently shown here that $\sin\left(\dfrac{\pi}{10}\right) = \dfrac{-1+ \sqrt{5}}{4}$.