Review question

# Can we simplify $\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6807

## Solution

1. Prove that $\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta = \cos 2\theta \sin \theta.$

#### Approach 1

We know $\sin 2\theta=2\sin\theta\cos\theta$, so the left hand side becomes \begin{align*} \cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta &{}= \cos 3\theta (2\sin\theta\cos\theta) - \cos 4\theta \sin \theta\\ &{}= \sin\theta(2\cos 3\theta\cos\theta - \cos 4\theta) \end{align*} Now $\cos3\theta\cos\theta$ appears in the expansion of $\cos 4\theta=\cos(3\theta+\theta)$, giving \begin{align*} &\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta\\ &\qquad{}=\sin\theta(2\cos 3\theta\cos\theta - \cos 4\theta)\\ &\qquad{}=\sin\theta(2\cos 3\theta \cos\theta - (\cos 3\theta\cos\theta- \sin3\theta\sin\theta))\\ &\qquad{}=\sin\theta(\cos 3\theta \cos\theta + \sin3\theta\sin\theta)\\ &\qquad{}=\sin\theta\cos(3\theta-\theta)\\ &\qquad{}=\sin\theta\cos 2\theta. \end{align*}

#### Approach 2

Using the trig expansions we know for $\sin(P+Q), \sin(P-Q), \cos(P+Q)$ and $\cos(P-Q)$, we have

\begin{align*} \sin P\cos Q &{}= \dfrac{1}{2}(\sin(P+Q)+\sin(P-Q))\\ \cos P\sin Q &{}= \dfrac{1}{2}(\sin(P+Q)-\sin(P-Q))\\ \cos P\cos Q &{}= \dfrac{1}{2}(\cos(P+Q)+\cos(P-Q))\\ \sin P\sin Q &{}= -\dfrac{1}{2}(\cos(P+Q)-\cos(P-Q)) \end{align*}

Now putting $A = P+Q$ and $B = P-Q$, we have

\begin{align*} \sin A+\sin B &{}= 2\sin \frac{A+B}{2}\cos \frac{A-B}{2} \\ \sin A-\sin B &{}= 2\cos \frac{A+B}{2}\sin \frac{A-B}{2} \\ \cos A+\cos B &{}= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2} \\ \cos A-\cos B &{}= -2\sin \frac{A+B}{2}\sin \frac{A-B}{2} \end{align*}
This gives \begin{align*} \cos 3\theta \sin 2\theta &{}- \cos 4\theta \sin \theta \\ &\quad{}=\dfrac{1}{2}(\sin 5\theta-\sin\theta-\sin 5\theta+\sin 3\theta)\\ &\quad{}=\dfrac{1}{2}(\sin 3\theta-\sin\theta)\\ &\quad{}=\cos 2\theta \sin \theta. \end{align*}
1. Starting from the relations \begin{align*} \sin(A+B) &{}= \sin A \cos B + \cos A \sin B,\\ \cos(A+B) &{}= \cos A \cos B - \sin A \sin B, \end{align*} obtain an expression for $\tan(A+B)$ in terms of $\tan A$ and $\tan B$.

We know $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, giving

\begin{align*} \tan(A+B) &{}= \frac{\sin(A+B)}{\cos(A+B)} \\ &{}= \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}. \end{align*}

Now we divide the numerator and denominator of the fraction by $\cos A\cos B$, giving

\begin{align*} \tan(A+B) &{}= \dfrac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A\sin B}{\cos A\cos B}} \\ &{}= \dfrac{\tan A + \tan B}{1 - \tan A\tan B} \end{align*}

Without the use of a calculator or tables, show that the sum of the three acute angles whose tangents are $\dfrac{2}{9}$, $\dfrac{1}{4}$ and $\dfrac{1}{3}$ is $\dfrac{\pi}{4}$.

Call the three angles $A$, $B$ and $C$, so that $\tan A = \dfrac{2}{9}$, $\tan B = \dfrac{1}{4}$ and $\tan C = \dfrac{1}{3}$.

Then \begin{align*} \tan(A+C) &{}= \frac{\tan A + \tan C}{1 - \tan A\tan C}\\ &= \frac{\frac{2}{9}+\frac{1}{3}}{1 - \frac{2}{9}\frac{1}{3}}\\ &= \frac{(\frac{5}{9})}{(\frac{25}{27})}\\ &= \frac{3}{5}. \end{align*} Now \begin{align*} \tan(A+C+B) &{}= \frac{\tan(A+C) + \tan B}{1 - \tan(A+C)\tan B}\\ &= \frac{\frac{3}{5}+\frac{1}{4}}{1 - \frac{3}{5}\frac{1}{4}}\\ &= \frac{(\frac{17}{20})}{(\frac{17}{20})}\\ &= 1. \end{align*}

We are told that $A$, $B$ and $C$ are acute. Also, since $\tan A$, $\tan B$ and $\tan C$ are all less than $1$, the angles $A$, $B$ and $C$ are all between $0$ and $\dfrac{\pi}{4}$. Therefore,$0 < A + B + C < \dfrac{3\pi}{4}$.

So the only possible solution of $\tan (A+B+C) = 1$ with $A+B+C$ in this range is $A+B+C = \dfrac{\pi}{4}$, as required.