Complementary angles

Two angles are complementary if they sum to $90^\circ$. In a right triangle, the two acute angles are always complementary because $A + B + 90^\circ = 180^\circ \Rightarrow A + B = 90^\circ$. This forces a beautiful pairing of trig ratios.

The identities

For any $\theta$ with $0^\circ < \theta < 90^\circ$: $\sin(90^\circ - \theta) = \cos\theta, \quad \cos(90^\circ - \theta) = \sin\theta,$ $\tan(90^\circ - \theta) = \cot\theta, \quad \cot(90^\circ - \theta) = \tan\theta,$ $\sec(90^\circ - \theta) = \csc\theta, \quad \csc(90^\circ - \theta) = \sec\theta.$

Why they hold

In a right triangle with acute angles $\theta$ and $90^\circ - \theta$, what is the opposite of $\theta$ is the adjacent of $90^\circ - \theta$, and vice versa. The hypotenuse is the same.

So:

• $\sin\theta = \text{opp}_\theta / \text{hyp} = \text{adj}_{90^\circ - \theta}/\text{hyp} = \cos(90^\circ - \theta)$.
• $\cos\theta = \text{adj}_\theta / \text{hyp} = \text{opp}_{90^\circ - \theta}/\text{hyp} = \sin(90^\circ - \theta)$.

Similar swaps give the other four identities.

Where they're useful

Simplification. A trig expression containing $\sin(90^\circ - \theta)$ becomes simpler if rewritten as $\cos\theta$. Long expressions often collapse to a number after applying complementary identities.

"Angle in disguise" problems. If you see $\sin 50^\circ$, note that $50^\circ = 90^\circ - 40^\circ$, so $\sin 50^\circ = \cos 40^\circ$. This lets you compare angles whose ratios aren't immediately on the standard list.

Combining with the Pythagorean identity. A common trick: replace $\sin 35^\circ$ with $\cos 55^\circ$ and then group with another $\cos 55^\circ$ to simplify.

Worked examples

Example 1. Simplify $\sin 65^\circ / \cos 25^\circ$.

$\cos 25^\circ = \sin (90^\circ - 25^\circ) = \sin 65^\circ$. So the ratio is $1$.

Example 2. Evaluate $\sin 18^\circ / \cos 72^\circ + \sec 32^\circ / \csc 58^\circ$.

$\cos 72^\circ = \sin 18^\circ$, so first ratio is $1$. $\csc 58^\circ = \sec 32^\circ$, so second ratio is $1$. Total $= 2$.

Example 3. If $\sin 3 A = \cos (A - 26^\circ)$, where $3 A$ is acute, find $A$.

Use $\cos\theta = \sin(90^\circ - \theta)$: $\sin 3 A = \sin(90^\circ - (A - 26^\circ)) = \sin(116^\circ - A)$.

So $3 A = 116^\circ - A \Rightarrow 4 A = 116^\circ \Rightarrow A = 29^\circ$.

Example 4. Show $\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ = 1$.

Pair: $\tan k^\circ \cdot \tan(90^\circ - k^\circ) = \tan k^\circ \cdot \cot k^\circ = 1$.

Pairs are $(1, 89), (2, 88), \ldots, (44, 46)$. That's $44$ pairs, all multiplying to $1$. The middle term is $\tan 45^\circ = 1$.

Product $= 1 \cdot 1 \cdot 1 \cdots 1 \cdot 1 = 1$. $\blacksquare$

Example 5. Evaluate $\cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ$.

$\sin 52^\circ = \cos 38^\circ$ and $\cos 52^\circ = \sin 38^\circ$.

Expression $= \cos 38^\circ \cdot \sin 38^\circ - \sin 38^\circ \cdot \cos 38^\circ = 0$.

Try it yourself

1. Simplify $\cos 27^\circ / \sin 63^\circ$.
2. Simplify $\tan 15^\circ \cdot \tan 75^\circ$.
3. Show $\sin 36^\circ - \cos 54^\circ = 0$.
4. If $\sin 5 A = \cos (4 A - 12^\circ)$, find $A$.
5. Evaluate $\dfrac{\sin 47^\circ}{\cos 43^\circ} + \dfrac{\cos 47^\circ}{\sin 43^\circ} - 4 \cos^2 45^\circ$.
6. If $\sec 4 A = \csc(A - 20^\circ)$, find $A$.
7. Evaluate $\sec 70^\circ \sin 20^\circ + \cos 20^\circ \csc 70^\circ$.
8. Show $\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ = 1$.
9. Express $\sin 67^\circ + \cos 75^\circ$ in terms of trig ratios of angles between $0^\circ$ and $45^\circ$.
10. Evaluate $\sin 30^\circ + \cos 60^\circ - \tan 45^\circ$.

Pitfalls / Insight

• The pair is $\theta$ and $90^\circ - \theta$, not $\theta$ and $-\theta$. Don't mix complementary with negation.
• $\sin$ and $\cos$ swap. $\tan$ and $\cot$ swap. $\sec$ and $\csc$ swap.
• For unusual angles like $50^\circ, 65^\circ$, complementary identities convert them into recognisable forms.

Insight. Complementary identities are a symmetry of the right triangle: each acute angle is the other's complement, and trig ratios pair up neatly.