Mathematics 9709 · Pure Mathematics 1

Inverse trigonometric functions

Pressing $\sin^{-1}$ asks: “which angle has this sine?”
But many angles have the same sine. There is no end to them. So the calculator must pick just one. This topic is about knowing which one it picks.

Each inverse always gives back an angle in one fixed range. This is called the principal value. The exam often tests the edges of these ranges. So they are worth more marks than they look.

Why we have to cut down the angle

$\sin x=\tfrac12$ at $30^\circ$ , $150^\circ$ , $390^\circ$ and many more. But a function may give only one output for each input.
So we cut down the domain to a part where the curve goes up once through every value. Then the inverse can only give back one fixed range. This range is the principal value.

The three principal-value ranges

Learn these three. The exam tests the edges of them:

$\sin^{-1}x$ gives an angle in $-90^\circ\le \theta\le 90^\circ$ (i.e. $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ ), input $-1\le x\le1$ .
$\cos^{-1}x$ gives an angle in $0\le \theta\le 180^\circ$ (i.e. $0$ to $\pi$ ), input $-1\le x\le1$ .
$\tan^{-1}x$ gives an angle in $-90^\circ< \theta< 90^\circ$ , and it accepts any input $x$ .

Quick check: $\sin^{-1}$ and $\tan^{-1}$ can give negative angles.
$\cos^{-1}$ never does. Its answers go from $0$ to $180^\circ$ only.

Each inverse turns a number into one angle: sin⁻¹ into [−90°, 90°], cos⁻¹ into [0°, 180°], and tan⁻¹ into (−90°, 90°) for any input.

Worked examples

Worked example

Find, in degrees, $\sin^{-1}\big(\tfrac12\big)$ , $\cos^{-1}\big(-\tfrac12\big)$ and $\tan^{-1}(1)$

Each one asks for the angle in that function's own range:

$\sin^{-1}\big(\tfrac12\big)=30^\circ$ (in $[-90^\circ,90^\circ]$ )

$\cos^{-1}\big(-\tfrac12\big)=120^\circ$ (in $[0,180^\circ]$ , must be obtuse, not $-60^\circ$ )

Answer

30^\circ,\quad 120^\circ,\quad 45^\circ

Worked example

$f(x)=3\sin\big(\tfrac{x}{2}\big)-1$ for $-\pi \le x \le \pi$ : find $f^{-1}(x)$

This cut-down domain makes $f$ one-one, so an inverse exists. To find it, set $y=3\sin\big(\tfrac{x}{2}\big)-1$ , swap $x$ and $y$ , then undo each step in turn:

Swap and isolate the sine: $x=3\sin\big(\tfrac{y}{2}\big)-1 \;\Rightarrow\; \tfrac{x+1}{3}=\sin\big(\tfrac{y}{2}\big)$ .
Apply $\sin^{-1}$ to undo the sine: $\tfrac{y}{2}=\sin^{-1}\big(\tfrac{x+1}{3}\big)$ .
Multiply by 2 and rename as $f^{-1}$ .

Answer

f^{-1}(x)=2\sin^{-1}\!\Big(\tfrac{x+1}{3}\Big)

Where the marks go

Common mistake

\cos^{-1}

of a negative number is obtuse, not negative.

\cos^{-1}(-\tfrac12)=120^\circ

, never

-60^\circ

Common mistake

To prove a function has an inverse, you must say it is one-one on the given domain. The domain was cut down on purpose. That is the reason it works. (9709-style)

Now you try

Find the value of $x$ satisfying $\sin^{-1}(x-1)=\tan^{-1}(3)$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Both sides are the same angle $\theta$ with $\tan\theta=3$ . Draw a right triangle. The opposite side is 3, the next side is 1, and the longest side is $\sqrt{10}$ . So $\sin\theta=\tfrac{3}{\sqrt{10}}$ .

The left side says $x-1=\sin\theta$ :

Answer

x=1+\tfrac{3}{\sqrt{10}}\approx 1.95

PreviousGraphs of sin, cos and tan NextTrigonometric identities

Inverse trigonometric functions

Pressing $\sin^{-1}$ asks: “which angle has this sine?”
But many angles have the same sine. There is no end to them. So the calculator must pick just one. This topic is about knowing which one it picks.

Each inverse always gives back an angle in one fixed range. This is called the principal value. The exam often tests the edges of these ranges. So they are worth more marks than they look.

Why we have to cut down the angle

$\sin x=\tfrac12$ at $30^\circ$ , $150^\circ$ , $390^\circ$ and many more. But a function may give only one output for each input.
So we cut down the domain to a part where the curve goes up once through every value. Then the inverse can only give back one fixed range. This range is the principal value.

The three principal-value ranges

Learn these three. The exam tests the edges of them:

$\sin^{-1}x$ gives an angle in $-90^\circ\le \theta\le 90^\circ$ (i.e. $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ ), input $-1\le x\le1$ .
$\cos^{-1}x$ gives an angle in $0\le \theta\le 180^\circ$ (i.e. $0$ to $\pi$ ), input $-1\le x\le1$ .
$\tan^{-1}x$ gives an angle in $-90^\circ< \theta< 90^\circ$ , and it accepts any input $x$ .

Quick check: $\sin^{-1}$ and $\tan^{-1}$ can give negative angles.
$\cos^{-1}$ never does. Its answers go from $0$ to $180^\circ$ only.

Each inverse turns a number into one angle: sin⁻¹ into [−90°, 90°], cos⁻¹ into [0°, 180°], and tan⁻¹ into (−90°, 90°) for any input.

Worked examples

Worked example

Find, in degrees, $\sin^{-1}\big(\tfrac12\big)$ , $\cos^{-1}\big(-\tfrac12\big)$ and $\tan^{-1}(1)$

Each one asks for the angle in that function's own range:

$\sin^{-1}\big(\tfrac12\big)=30^\circ$ (in $[-90^\circ,90^\circ]$ )

$\cos^{-1}\big(-\tfrac12\big)=120^\circ$ (in $[0,180^\circ]$ , must be obtuse, not $-60^\circ$ )

Answer

30^\circ,\quad 120^\circ,\quad 45^\circ

Worked example

$f(x)=3\sin\big(\tfrac{x}{2}\big)-1$ for $-\pi \le x \le \pi$ : find $f^{-1}(x)$

This cut-down domain makes $f$ one-one, so an inverse exists. To find it, set $y=3\sin\big(\tfrac{x}{2}\big)-1$ , swap $x$ and $y$ , then undo each step in turn:

Swap and isolate the sine: $x=3\sin\big(\tfrac{y}{2}\big)-1 \;\Rightarrow\; \tfrac{x+1}{3}=\sin\big(\tfrac{y}{2}\big)$ .
Apply $\sin^{-1}$ to undo the sine: $\tfrac{y}{2}=\sin^{-1}\big(\tfrac{x+1}{3}\big)$ .
Multiply by 2 and rename as $f^{-1}$ .

Answer

f^{-1}(x)=2\sin^{-1}\!\Big(\tfrac{x+1}{3}\Big)

Where the marks go

Common mistake

\cos^{-1}

of a negative number is obtuse, not negative.

\cos^{-1}(-\tfrac12)=120^\circ

, never

-60^\circ

Common mistake

To prove a function has an inverse, you must say it is one-one on the given domain. The domain was cut down on purpose. That is the reason it works. (9709-style)

Now you try

Find the value of $x$ satisfying $\sin^{-1}(x-1)=\tan^{-1}(3)$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

The left side says $x-1=\sin\theta$ :

Answer

x=1+\tfrac{3}{\sqrt{10}}\approx 1.95