Practice questions

Write everything with sines and cosines, and use the common denominator ${\cos x}$:

$\begin{align*}&\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}\cdot\sin x\\&=\dfrac{1-\sin^2 x}{\cos x}\\&=\dfrac{\cos^2 x}{\cos x}=\cos x\end{align*}$

Combine the fractions on the left. The common denominator is ${\sin x\,(1+\cos x)}$:

$\dfrac{\sin^2 x+(1+\cos x)^2}{\sin x\,(1+\cos x)}$

Expand the square on top:

$\begin{align*}&(1+\cos x)^2\\&=1+2\cos x+\cos^2 x\end{align*}$

On top, ${\sin^2 x+\cos^2 x=1}$, so the top is ${2+2\cos x=2(1+\cos x)}$. Cancel ${1+\cos x}$:

$\begin{align*}&\dfrac{2(1+\cos x)}{\sin x\,(1+\cos x)}\\&=\dfrac{2}{\sin x}=2\csc x\end{align*}$

Start with the left side:

$\begin{align*}&\tan x+\cot x\\&=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}\\&=\dfrac{\sin^2 x+\cos^2 x}{\sin x\cos x}\\&=\dfrac{1}{\sin x\cos x}\\&=\dfrac{1}{\cos x}\cdot\dfrac{1}{\sin x}=\sec x\csc x\end{align*}$

One value of ${x}$ where it fails is enough. Take ${x=\dfrac{\pi}{4}}$:

$\begin{align*}&\sin\dfrac{\pi}{4}+\cos\dfrac{\pi}{4}\\&=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}=\sqrt{2}\end{align*}$

This is not ${1}$, so the equation is not an identity.

Write ${105^\circ=60^\circ+45^\circ}$:

$\begin{align*}&\sin 105^\circ\\&=\sin 60^\circ\cos 45^\circ+\cos 60^\circ\sin 45^\circ\\&=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}\\&=\dfrac{\sqrt{6}+\sqrt{2}}{4}\end{align*}$

Write ${15^\circ=45^\circ-30^\circ}$, with ${\tan 45^\circ=1}$ and ${\tan 30^\circ=\dfrac{1}{\sqrt{3}}}$:

$\tan 15^\circ=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

(We multiplied the top and bottom by ${\sqrt{3}}$.) Now multiply by the conjugate ${\sqrt{3}-1}$:

$\begin{align*}&\dfrac{(\sqrt{3}-1)^2}{3-1}\\&=\dfrac{3-2\sqrt{3}+1}{2}=2-\sqrt{3}\end{align*}$

Find the missing values. Sine is negative in quadrant IV, and cosine is negative in quadrant II:

$\sin A=-\sqrt{1-\dfrac{144}{169}}=-\dfrac{5}{13}$

$\cos B=-\sqrt{1-\dfrac{16}{25}}=-\dfrac{3}{5}$

Now use the formula:

$\begin{align*}&\cos(A-B)\\&=\cos A\cos B+\sin A\sin B\\&=\dfrac{12}{13}\left(-\dfrac{3}{5}\right)+\left(-\dfrac{5}{13}\right)\dfrac{4}{5}\\&=-\dfrac{36}{65}-\dfrac{20}{65}=-\dfrac{56}{65}\end{align*}$

Here ${a=\sqrt{3}}$ and ${b=-1}$, so ${k=\sqrt{3+1}=2}$.

The angle ${\varphi}$ has ${\cos\varphi=\dfrac{\sqrt{3}}{2}}$ and ${\sin\varphi=-\dfrac{1}{2}}$. That is ${\varphi=-\dfrac{\pi}{6}}$. So

$\sqrt{3}\sin x-\cos x=2\sin\left(x-\dfrac{\pi}{6}\right)$

Since ${\tan x=\dfrac{y}{x}=\dfrac{-3}{-4}}$ in quadrant III, take the point ${(-4,-3)}$, with ${r=5}$. So ${\sin x=-\dfrac{3}{5}}$ and ${\cos x=-\dfrac{4}{5}}$.

$\sin 2x=2\left(-\dfrac{3}{5}\right)\left(-\dfrac{4}{5}\right)=\dfrac{24}{25}$

$\cos 2x=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}$

${22.5^\circ}$ is half of ${45^\circ}$, in quadrant I, so use ${+}$:

$\begin{align*}\cos 22.5^\circ&=\sqrt{\dfrac{1+\cos 45^\circ}{2}}\\&=\sqrt{\dfrac{1+\dfrac{\sqrt{2}}{2}}{2}}\\&=\sqrt{\dfrac{2+\sqrt{2}}{4}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\end{align*}$

Expand the left side, and regroup:

$\begin{align*}&(\sin x+\cos x)^2\\&=\sin^2 x+2\sin x\cos x+\cos^2 x\\&=(\sin^2 x+\cos^2 x)+2\sin x\cos x\\&=1+\sin 2x\end{align*}$

Write ${\cos^4 x=(\cos^2 x)^2}$, and use ${\cos^2 x=\dfrac{1+\cos 2x}{2}}$:

$\begin{align*}&\cos^4 x\\&=\left(\dfrac{1+\cos 2x}{2}\right)^2\\&=\dfrac{1+2\cos 2x+\cos^2 2x}{4}\end{align*}$

Use the same formula again: ${\cos^2 2x=\dfrac{1+\cos 4x}{2}}$. The top becomes ${\dfrac{3}{2}+2\cos 2x+\dfrac{1}{2}\cos 4x}$. Multiply the top and the bottom by ${2}$:

${\cos^4 x=\dfrac{3+4\cos 2x+\cos 4x}{8}}$

First, ${\cos u}$ is positive in quadrant IV:

$\cos u=\sqrt{1-\dfrac{25}{169}}=\dfrac{12}{13}$

Dividing ${\dfrac{3\pi}{2}<u<2\pi}$ by ${2}$ gives ${\dfrac{3\pi}{4}<\dfrac{u}{2}<\pi}$. So ${\dfrac{u}{2}}$ is in quadrant II, where sine is positive:

$\begin{align*}\sin\dfrac{u}{2}&=\sqrt{\dfrac{1-\dfrac{12}{13}}{2}}=\sqrt{\dfrac{1}{26}}\\&=\dfrac{1}{\sqrt{26}}=\dfrac{\sqrt{26}}{26}\end{align*}$

Use the formula for ${\cos A\cos B}$ with ${A=4x}$ and ${B=2x}$:

$\begin{align*}&\cos 4x\cos 2x\\&=\dfrac{1}{2}(\cos 6x+\cos 2x)\end{align*}$

Use the formula for ${\sin A\sin B}$:

$\begin{align*}&\sin 75^\circ\sin 15^\circ\\&=\dfrac{1}{2}(\cos 60^\circ-\cos 90^\circ)\\&=\dfrac{1}{2}\left(\dfrac{1}{2}-0\right)=\dfrac{1}{4}\end{align*}$

The average is ${4x}$, and half the difference is ${x}$. Remember the minus sign:

${\cos 5x-\cos 3x=-2\sin 4x\sin x}$

The average is ${45^\circ}$, and half the difference is ${30^\circ}$:

$\begin{align*}&\sin 75^\circ-\sin 15^\circ\\&=2\cos 45^\circ\sin 30^\circ\\&=2\cdot\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}}{2}\end{align*}$