Practice questions

The angle in ${[0,\pi]}$ with cosine ${\dfrac{\sqrt{3}}{2}}$ is ${\dfrac{\pi}{6}}$. So

$\cos^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}$

The angle in $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ with sine ${-1}$ is ${-\dfrac{\pi}{2}}$. So ${\sin^{-1}(-1)=-\dfrac{\pi}{2}}$.

Since $\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$, we get

$\tan^{-1}\dfrac{\sqrt{3}}{3}=\dfrac{\pi}{6}$

${\dfrac{5\pi}{3}}$ is not in ${[0,\pi]}$, so the functions do not cancel. Work from the inside:

${\cos\dfrac{5\pi}{3}=\dfrac{1}{2}}$

Then

${\cos^{-1}\dfrac{1}{2}=\dfrac{\pi}{3}}$

Let ${\theta=\tan^{-1}\dfrac{4}{3}}$, so ${\tan\theta=\dfrac{4}{3}}$ and ${\theta}$ is in quadrant I. Draw a right triangle with opposite side ${4}$ and adjacent side ${3}$. The hypotenuse is ${\sqrt{16+9}=5}$. So

$\sin\left(\tan^{-1}\dfrac{4}{3}\right)=\sin\theta=\dfrac{4}{5}$

Let ${\theta=\sin^{-1}x}$, so ${\sin\theta=x}$ and ${-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}}$. On this interval, ${\cos\theta\ge 0}$. By the Pythagorean identity,

${\cos(\sin^{-1}x)=\cos\theta=\sqrt{1-x^2}}$

Get sine alone: ${\sin x=-\dfrac{\sqrt{2}}{2}}$. Sine is negative in quadrants III and IV, with reference angle ${\dfrac{\pi}{4}}$:

${x=\dfrac{5\pi}{4}+2k\pi}$or${x=\dfrac{7\pi}{4}+2k\pi}$

Get tangent alone: ${\tan x=\dfrac{1}{\sqrt{3}}}$. In $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ that is ${x=\dfrac{\pi}{6}}$. Tangent has period ${\pi}$:

${x=\dfrac{\pi}{6}+k\pi}$

Cosine is zero at ${\dfrac{\pi}{2}}$ plus multiples of ${\pi}$:

${3x=\dfrac{\pi}{2}+k\pi}$, so ${x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}}$

Take ${k=0,1,2,3,4,5}$ to stay in ${[0,2\pi)}$:

${x=\dfrac{\pi}{6}}$, ${\dfrac{\pi}{2}}$, ${\dfrac{5\pi}{6}}$, ${\dfrac{7\pi}{6}}$, ${\dfrac{3\pi}{2}}$, ${\dfrac{11\pi}{6}}$

In radian mode, ${\cos^{-1}(-0.4)\approx 1.9823}$. That is in quadrant II. Cosine is also negative in quadrant III. By symmetry, the other solution is ${2\pi}$ minus the first:

${x\approx 2\pi-1.9823\approx 4.3009}$

The solutions are ${x\approx 1.9823}$ and ${x\approx 4.3009}$.

Factor out ${\cos x}$:

${\cos x\,(2\cos x-1)=0}$

• ${\cos x=0}$: ${x=\dfrac{\pi}{2}}$ or ${\dfrac{3\pi}{2}}$.
• ${\cos x=\dfrac{1}{2}}$: ${x=\dfrac{\pi}{3}}$ or ${\dfrac{5\pi}{3}}$.

Replace ${\sin^2 x}$ by ${1-\cos^2 x}$:

${2(1-\cos^2 x)+3\cos x-3=0}$

Multiply out, and multiply by ${-1}$:

$\begin{align*}2\cos^2 x-3\cos x+1&=0\\(2\cos x-1)(\cos x-1)&=0\end{align*}$

${\cos x=\dfrac{1}{2}}$ gives ${\dfrac{\pi}{3}}$ and ${\dfrac{5\pi}{3}}$. ${\cos x=1}$ gives ${0}$. The solutions are ${0}$, ${\dfrac{\pi}{3}}$, ${\dfrac{5\pi}{3}}$.

Use ${\cos 2x=1-2\sin^2 x}$, so only sine appears:

$\begin{align*}1-2\sin^2 x&=\sin x\\2\sin^2 x+\sin x-1&=0\\(2\sin x-1)(\sin x+1)&=0\end{align*}$

${\sin x=\dfrac{1}{2}}$ gives ${\dfrac{\pi}{6}}$ and ${\dfrac{5\pi}{6}}$. ${\sin x=-1}$ gives ${\dfrac{3\pi}{2}}$.

Use ${\sin 2x=2\sin x\cos x}$, and factor out ${\sin x}$:

$\begin{align*}2\sin x\cos x-\sin x&=0\\\sin x\,(2\cos x-1)&=0\end{align*}$

${\sin x=0}$ gives ${0}$ and ${\pi}$. ${\cos x=\dfrac{1}{2}}$ gives ${\dfrac{\pi}{3}}$ and ${\dfrac{5\pi}{3}}$.

Square both sides. The left side becomes

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

So

$\begin{align*}1-2\sin x\cos x&=1\\\sin x\cos x&=0\end{align*}$

The candidates are ${0}$, ${\dfrac{\pi}{2}}$, ${\pi}$, ${\dfrac{3\pi}{2}}$. Check each in ${\cos x-\sin x=1}$:

• ${x=0}$: ${1-0=1}$. Yes.
• ${x=\dfrac{\pi}{2}}$: ${0-1=-1}$. No.
• ${x=\pi}$: ${-1-0=-1}$. No.
• ${x=\dfrac{3\pi}{2}}$: ${0-(-1)=1}$. Yes.

The solutions are ${0}$ and ${\dfrac{3\pi}{2}}$.