Arccos

Idea

Cosine cannot share arcsin's window. Between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$ it rises then falls, so it hits most values twice: $\cos\left(-\dfrac{\pi}{3}\right)$ and $\cos\dfrac{\pi}{3}$ are both $\dfrac{1}{2}$ .

So cosine gets its own window, $[0, \pi]$ . There it falls steadily from $1$ to $-1$ and never repeats.

Rule

$y = \arccos x \iff \cos y = x$

$D = [-1, 1] \qquad R = [0, \pi]$

Arccos turns a number into an angle, and that angle only ever lands in the first or second quadrant.

$\arccos(\text{positive}) \rightarrow$ first quadrant

$\arccos(\text{negative}) \rightarrow$ second quadrant

The unit circle

Try this. Try 0°, 90°, 180° and 270°. The horizontal projection is cosine; the vertical projection is sine. Track their signs between axes.

Angle (degrees)120

θ = 120° = 0.667π rad · cos θ = -0.5 · sin θ = 0.866 · arccos window: 0 to 180 degrees: inside

How it is used

1. A negative input reflects, it never flips sign

Students borrow arcsin's rule here. Arccos is not odd.

\arccos(-x) = \pi - \arccos x

A negative input gives an obtuse angle, never a negative one. This is the one rule in this lesson that arcsin does not prepare you for - the first worked example shows what goes wrong when you borrow.

2. What carries over from arcsin unchanged

Everything else you learned there transfers directly.

The round trip works the same way: $\cos(\arccos x) = x$ for every $x$ in $[-1, 1]$ .

The releasing pattern is unchanged. Arccos is what releases $x$ from inside a cosine.

No special angle? Draw the triangle, as in Arcsin, section 4 - starting from $\cos\alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ .

The domain is the same $[-1, 1]$ , and inputs outside it are just as undefined.

3. The values worth knowing

$x$	$-1$	$-\dfrac{\sqrt{3}}{2}$	$-\dfrac{\sqrt{2}}{2}$	$-\dfrac{1}{2}$	$0$	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	$1$
$\arccos x$	$180^\circ$	$150^\circ$	$135^\circ$	$120^\circ$	$90^\circ$	$60^\circ$	$45^\circ$	$30^\circ$	$0^\circ$

As $x$ grows, $\arccos x$ falls. Arcsin climbs. The negative half of this table is the positive half through the reflect rule, so learn the right side and the rule.

4. The domain of an arccos expression

Whatever sits inside must stay between $-1$ and $1$ . That one requirement fixes the domain, and solving it is a compound inequality.

-1 \le \text{inside} \le 1

5. Two sums worth recognising

Sum	Equals	Condition
$\arcsin x + \arccos x$	$\dfrac{\pi}{2}$	any $x$ in $[-1, 1]$
$\arccos x + \arccos y$	$\dfrac{\pi}{2}$	$x^2 + y^2 = 1$ , both positive

The second is a right triangle again: $x$ and $y$ are its two legs over one hypotenuse, so they name its two acute angles. Both must be positive, or the angles are not acute and the sum is something else.

Worked example - the rule arcsin lends does not work here

Evaluate $\sin\left(\arccos\left(-\dfrac{\sqrt{2}}{2}\right)\right)$ .

Remember

$\arccos\dfrac{\sqrt{2}}{2} = 45^\circ$ , from the values table sine is positive in the second quadrant

Here is the tempting wrong move. Arcsin is odd, so a negative input just flips the sign - borrow that:

\arccos\left(-\frac{\sqrt{2}}{2}\right) = -45^\circ \quad \text{(wrong)}

But $-45^\circ$ is not in $[0, \pi]$ . Arccos can never answer with a negative angle, so the borrowed rule has put us outside the range. Reflect instead:

\arccos\left(-\frac{\sqrt{2}}{2}\right) = 180^\circ - 45^\circ = 135^\circ

The question is now $\sin 135^\circ$ , a second-quadrant angle, where sine is still positive.

\sin 135^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}

Compare the two paths. The wrong one gives $\sin(-45^\circ) = -\dfrac{\sqrt{2}}{2}$ : right size, wrong sign. That sign is exactly what the reflect rule protects.

Worked example - a ratio with no special angle

Evaluate $\tan\left(\arccos\dfrac{3}{7}\right)$ .

Remember

no special angle? Name it and draw the triangle - Arcsin, section 4 $\cos\alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

Name the angle. There is no special angle here, so do not look for one.

\arccos\frac{3}{7} = \alpha \quad \text{so} \quad \cos\alpha = \frac{3}{7}

Cosine is adjacent over hypotenuse, so the triangle has adjacent $3$ and hypotenuse $7$ .
Pythagoras supplies the opposite side.

\sqrt{7^2 - 3^2} = \sqrt{40} = 2\sqrt{10}

Now read the tangent straight off the completed triangle.

\tan\alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{2\sqrt{10}}{3}

The angle itself was never needed. That is the whole point of the method.

A right triangle

Try this. Compare the labeled lengths with sine, cosine and tangent. If size is adjustable, scale the triangle: lengths change together, but the ratios stay fixed.

θ = 64.623° · sin θ = 0.904 · cos θ = 0.429 · tan θ = 2.108

Explore

Drag the probe and read both values. Add them: the total is always $\pi$ , about $3.14$ - the readout under the graph keeps the running sum. As one angle falls the other rises to keep it.

A negative input never gives a negative angle. It gives the obtuse partner. Compare this with arcsin's picture in the last lesson, where the two readings added to zero instead.

Inverse trigonometric functions

Try this. Move the probe from positive x to negative x. Compare the two outputs in radians and the indicated symmetry or sum.

Probe x0.5

arccos(0.5) = 1.047 rad · arccos(-0.5) = 2.094 rad · Sum = 3.142 rad