Math Core

You know the definitions of arcsine, arccosine and arctangent. Now you will evaluate them: exactly, using the unit circle values you already know, and approximately, with a calculator. The one skill that ties it all together is keeping every answer inside the right range.

A two-step method

Every exact evaluation follows the same pattern.

Evaluating an inverse trig function

Find the reference angle. Ignore the sign of the input and ask which first-quadrant angle has that ratio. For special values this is $\dfrac{\pi}{6}$ , $\dfrac{\pi}{4}$ or $\dfrac{\pi}{3}$ (or $0$ or $\dfrac{\pi}{2}$ ).
Place it in the correct range. A positive input gives the first-quadrant angle itself. A negative input sends the answer to a second quadrant, which depends on the function:
- $\arcsin$ and $\arctan$ : Quadrant IV, written as a negative angle $-\theta$ .
- $\arccos$ : Quadrant II, the angle $\pi - \theta$ .

Here are the first-quadrant values you need for step 1:

angle	$\sin$	$\cos$	$\tan$
$0$	$0$	$1$	$0$
$\dfrac{\pi}{6}$ ( $30^\circ$ )	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{3}}{3}$
$\dfrac{\pi}{4}$ ( $45^\circ$ )	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$
$\dfrac{\pi}{3}$ ( $60^\circ$ )	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$
$\dfrac{\pi}{2}$ ( $90^\circ$ )	$1$	$0$	undefined

Remember that $\dfrac{\sqrt{3}}{3}$ and $\dfrac{1}{\sqrt{3}}$ are the same number, so either form can appear in a problem.

Positive inputs

With a positive input, the answer is just the first-quadrant angle, since Quadrant I is part of every range.

Worked example: Positive inputs

Find the exact value of each, in radians.

$\arcsin\dfrac{\sqrt{2}}{2}$
$\arccos\dfrac{\sqrt{3}}{2}$
$\arctan\sqrt{3}$

Solutions.

$\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$ , so $\arcsin\dfrac{\sqrt{2}}{2} = \dfrac{\pi}{4}$ .
$\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$ , so $\arccos\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6}$ .
$\tan\dfrac{\pi}{3} = \sqrt{3}$ , so $\arctan\sqrt{3} = \dfrac{\pi}{3}$ .

Negative inputs

Negative inputs are where the ranges do real work. Picture the unit circle:

Arcsine and arctangent live on the right half of the circle, from $-\dfrac{\pi}{2}$ up to $\dfrac{\pi}{2}$ . A negative sine (a point below the $x$ -axis) or a negative tangent puts the answer in Quadrant IV, and you describe that angle by rotating clockwise, so it is negative.
Arccosine lives on the top half, from $0$ to $\pi$ . A negative cosine (a point left of the $y$ -axis) puts the answer in Quadrant II.

This gives three handy rules:

\arcsin(-x) = -\arcsin x, \qquad \arctan(-x) = -\arctan x, \qquad \arccos(-x) = \pi - \arccos x.

Worked example: Negative inputs

Find the exact value of each, in radians.

$\arcsin\left(-\dfrac{1}{2}\right)$
$\arccos\left(-\dfrac{\sqrt{2}}{2}\right)$
$\arctan(-1)$

Solutions.

The reference angle is $\dfrac{\pi}{6}$ , since $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$ . The input is negative, so the answer is in Quadrant IV: $\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$ .
The reference angle is $\dfrac{\pi}{4}$ . The input is negative, so the answer is in Quadrant II: $\arccos\left(-\dfrac{\sqrt{2}}{2}\right) = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$ .
The reference angle is $\dfrac{\pi}{4}$ , since $\tan\dfrac{\pi}{4} = 1$ . The answer is in Quadrant IV: $\arctan(-1) = -\dfrac{\pi}{4}$ .

Check each one by applying the original function: $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$ , $\cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2}$ , and $\tan\left(-\dfrac{\pi}{4}\right) = -1$ . Each angle is also inside the right range.

Common mistake

Don't write $\arcsin\left(-\dfrac{1}{2}\right) = \dfrac{7\pi}{6}$ or $\dfrac{11\pi}{6}$ . Both angles have sine $-\dfrac{1}{2}$ , but neither is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ . The same goes for arccosine: $\arccos\left(-\dfrac{\sqrt{2}}{2}\right)$ is $\dfrac{3\pi}{4}$ , never $-\dfrac{\pi}{4}$ or $\dfrac{5\pi}{4}$ . Arccosine is never negative. After you find an answer, always ask: "Is this angle in the range?"

Answers in degrees

If a problem asks for degrees, use the same method with degree ranges: arcsine and arctangent give angles from $-90^\circ$ to $90^\circ$ , and arccosine gives angles from $0^\circ$ to $180^\circ$ . For example, $\arccos(-1) = 180^\circ$ , $\;\arcsin\left(-\dfrac{\sqrt{2}}{2}\right) = -45^\circ$ , and $\arctan\left(-\sqrt{3}\right) = -60^\circ$ .

Using a calculator

Most inputs are not special values, and then you need a calculator. Check the mode first: radian mode gives radians and degree mode gives degrees. The calculator always returns the angle in the standard range, so it follows the same rules you do.

Worked example: Calculator values

Approximate each value.

$\arccos(-0.3)$ in radians, to the nearest hundredth.
$\arctan(-2.5)$ in degrees, to the nearest tenth.

Solutions.

In radian mode, $\cos^{-1}(-0.3) \approx 1.8755$ , so the answer is about $1.88$ . That is between $\dfrac{\pi}{2} \approx 1.57$ and $\pi \approx 3.14$ , a Quadrant II angle, as expected for a negative input to arccosine.
In degree mode, $\tan^{-1}(-2.5) \approx -68.199^\circ$ , so the answer is about $-68.2^\circ$ , a Quadrant IV angle.

The other three inverse functions

Your calculator probably has no $\sec^{-1}$ , $\csc^{-1}$ or $\cot^{-1}$ keys. You rarely need them, but you can handle them with reciprocals. Since $\sec\theta = \dfrac{1}{\cos\theta}$ , the angle whose secant is $x$ is the angle whose cosine is $\dfrac{1}{x}$ :

\operatorname{arcsec} x = \arccos\frac{1}{x}, \qquad \operatorname{arccsc} x = \arcsin\frac{1}{x} \qquad (\text{for } |x| \ge 1).

Worked example: An inverse secant

Find $\operatorname{arcsec} 2$ exactly.

Solution. $\operatorname{arcsec} 2 = \arccos\dfrac{1}{2} = \dfrac{\pi}{3}$ . Check: $\sec\dfrac{\pi}{3} = \dfrac{1}{\cos(\pi/3)} = \dfrac{1}{1/2} = 2$ .

Tip

Before you commit to an answer, do a quick sign check. Arccosine is always positive or zero. Arcsine and arctangent have the same sign as the input. If your answer breaks either rule, it is outside the range.