Chapter 7: Complementary Angle Formulas
Overview
Complementary angle formulas (quick reference)
\begin{align*} \sin\left(\tfrac{\pi}{2} - \theta\right) &= \cos\theta & \cos\left(\tfrac{\pi}{2} - \theta\right) &= \sin\theta \\ \tan\left(\tfrac{\pi}{2} - \theta\right) &= \cot\theta & \cot\left(\tfrac{\pi}{2} - \theta\right) &= \tan\theta \\ \sec\left(\tfrac{\pi}{2} - \theta\right) &= \csc\theta & \csc\left(\tfrac{\pi}{2} - \theta\right) &= \sec\theta \end{align*}The angles $\theta$ and $\dfrac{\pi}{2}-\theta$ are complementary (they sum to a right angle). Each trigonometric value swaps with its co- counterpart.
This page proves all six identities above using the definitions of the trigonometric functions, using a right-triangle picture, an explanation of the cofunction etymology, three worked examples, and four practice exercises.
Definitions of the trigonometric functions
Before the proofs, we recall the right-triangle definitions of the trigonometric functions.
Figure 1: Right triangle (adjacent $x$, opposite $y$, hypotenuse $r$, angle $\theta$).
With the adjacent leg of length $x$, the opposite leg of length $y$, the hypotenuse of length $r$, and angle $\theta$ at vertex A as shown in Figure 1, the trigonometric functions are defined by:
Definitions of the trigonometric functions
$$ \sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{y}{r} \tag{1}\label{def-sin} $$ $$ \cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{x}{r} \tag{2}\label{def-cos} $$ $$ \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{y}{x} = \dfrac{\sin\theta}{\cos\theta} \tag{3}\label{def-tan} $$ $$ \cot\theta = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{x}{y} = \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\tan\theta} \tag{4}\label{def-cot} $$The key observation is that $\tan\theta$ is “opposite over adjacent” while $\cot\theta$ is “adjacent over opposite” — the two are reciprocals.
Cofunction identities for $\sin, \cos$
Theorem
$$ \sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\theta, \quad \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta $$Proof. This is the basic result from which the $\tan$ identity in the next section follows.
Step 1: Relation between the two acute angles
Figure 2: The two acute angles $\theta$ and $\pi/2 - \theta$ in a right triangle.
In the triangle of Figure 2 the right angle is $\angle C = \dfrac{\pi}{2}$. Since the interior angles sum to $\pi$,
$$ \angle A + \angle B = \pi - \dfrac{\pi}{2} = \dfrac{\pi}{2}. $$So if we set $\angle A = \theta$, then $\angle B = \dfrac{\pi}{2} - \theta$.
Step 2: Trigonometric values at θ (∠A)
From angle $\theta$ (∠A), the opposite leg has length $y$, the adjacent leg has length $x$, and the hypotenuse has length $r$, so the definitions \eqref{def-sin}, \eqref{def-cos} give:
$$ \sin\theta = \dfrac{y}{r} \quad \text{(re-stating \eqref{def-sin})} $$ $$ \cos\theta = \dfrac{x}{r} \quad \text{(re-stating \eqref{def-cos})} $$Step 3: Trigonometric values at π/2 − θ (∠B)
Now we look at the triangle from angle $\dfrac{\pi}{2} - \theta$ (∠B). Re-drawing Figure 2 with ∠B at the lower left gives Figure 3.
Figure 3: Figure 2 redrawn so that ∠B (the angle $\pi/2 - \theta$) is at the lower left.
The hypotenuse is still $r$, but the opposite leg is now $x$ and the adjacent leg is now $y$. So from $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$ and $\cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ we obtain:
$$ \sin\left(\dfrac{\pi}{2} - \theta\right) = \dfrac{x}{r} \tag{5}\label{sincomp} $$ $$ \cos\left(\dfrac{\pi}{2} - \theta\right) = \dfrac{y}{r} \tag{6}\label{coscomp} $$Step 4: Comparison
Lining up the definitions \eqref{def-sin}, \eqref{def-cos} with the Step 3 results \eqref{sincomp}, \eqref{coscomp}:
| Function | From $\angle A\;(\theta)$ | From $\angle B\;\left(\dfrac{\pi}{2}-\theta\right)$ |
|---|---|---|
| $\sin$ | $\sin\theta = \dfrac{y}{r}$ \eqref{def-sin} | $\sin\!\left(\dfrac{\pi}{2}-\theta\right) = \dfrac{x}{r}$ \eqref{sincomp} |
| $\cos$ | $\cos\theta = \dfrac{x}{r}$ \eqref{def-cos} | $\cos\!\left(\dfrac{\pi}{2}-\theta\right) = \dfrac{y}{r}$ \eqref{coscomp} |
Reading the table diagonally:
- \eqref{sincomp} matches \eqref{def-cos}, i.e., $\sin\!\left(\dfrac{\pi}{2}-\theta\right) = \cos\theta$.
- \eqref{coscomp} matches \eqref{def-sin}, i.e., $\cos\!\left(\dfrac{\pi}{2}-\theta\right) = \sin\theta$.
Switching the reference angle simply swaps the opposite and adjacent legs — that is the reason behind the identities.
$\blacksquare$
Cofunction identities for $\tan, \cot$
Theorem
$$ \tan\left(\dfrac{\pi}{2} - \theta\right) = \cot\theta $$Proof.
Step 1: Apply the definition of tan
Using $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$ from \eqref{def-tan},
$$ \tan\left(\dfrac{\pi}{2} - \theta\right) = \dfrac{\sin\left(\dfrac{\pi}{2} - \theta\right)}{\cos\left(\dfrac{\pi}{2} - \theta\right)} \tag{7}\label{tancomp-expand} $$Step 2: Substitute the sin, cos cofunction identities
From the previous section,
$$ \sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\theta, \quad \cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta, $$which we substitute into \eqref{tancomp-expand}.
Step 3: Simplify
The substitution gives
$$ \tan\left(\dfrac{\pi}{2} - \theta\right) = \dfrac{\cos\theta}{\sin\theta} \tag{8}\label{tancomp-result} $$Step 4: Compare with the definition of cot
The definition \eqref{def-cot} gives $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$, which matches \eqref{tancomp-result}.
$\blacksquare$
Theorem
$$ \cot\left(\dfrac{\pi}{2} - \theta\right) = \tan\theta $$Proof. Using the definition \eqref{def-cot} of $\cot$ together with the cofunction identities for $\sin, \cos$:
$$ \cot\left(\dfrac{\pi}{2} - \theta\right) = \dfrac{\cos\left(\dfrac{\pi}{2} - \theta\right)}{\sin\left(\dfrac{\pi}{2} - \theta\right)} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta. $$$\blacksquare$
Worked examples
Example 1
Use the complementary angle formula to evaluate $\sin 60°$.
Solution
Since $60° = 90° - 30°$, applying $\sin(90° - \theta) = \cos\theta$ gives
$$ \sin 60° = \sin(90° - 30°) = \cos 30°. $$The standard value $\cos 30° = \dfrac{\sqrt{3}}{2}$ then yields
Example 2
Simplify the following expression.
$$ \cos\left(\dfrac{\pi}{2} - \theta\right) \cdot \dfrac{1}{\cos\theta} $$Solution
Substituting $\cos\!\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta$:
$$ \cos\left(\dfrac{\pi}{2} - \theta\right) \cdot \dfrac{1}{\cos\theta} = \sin\theta \cdot \dfrac{1}{\cos\theta} = \dfrac{\sin\theta}{\cos\theta}. $$Example 3
Express $\tan 70°$ using a trigonometric function other than $\tan$.
Solution
Since $70° = 90° - 20°$, applying $\tan(90° - \theta) = \cot\theta$ gives
$$ \tan 70° = \tan(90° - 20°) = \cot 20°. $$Using the definition $\cot\theta = \dfrac{1}{\tan\theta}$ this can also be written as
Exercises
Exercise 1
Use the complementary angle formula to verify $\sin 60° = \cos 30°$, and find the common value.
Hint
Note that $60° + 30° = 90°$, and apply $\sin(90° - \theta) = \cos\theta$ with $\theta = 30°$.
Solution
Setting $\theta = 30°$ gives $\sin(90° - 30°) = \sin 60°$.
The complementary angle formula $\sin(90° - \theta) = \cos\theta$ then yields
$$ \sin 60° = \cos 30°, $$and both equal $\dfrac{\sqrt{3}}{2}$.
Exercise 2
Use $\cos\!\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta$ to simplify
$$ \sin^2\theta + \cos^2\!\left(\dfrac{\pi}{2} - \theta\right). $$Hint
First apply the cofunction identity to $\cos\!\left(\dfrac{\pi}{2} - \theta\right)$, then simplify.
Solution
From $\cos\!\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta$,
$$ \cos^2\!\left(\dfrac{\pi}{2} - \theta\right) = \sin^2\theta. $$Therefore,
$$ \sin^2\theta + \cos^2\!\left(\dfrac{\pi}{2} - \theta\right) = \sin^2\theta + \sin^2\theta = 2\sin^2\theta. $$Exercise 3
Use $\tan(90° - \theta) = \cot\theta$ to evaluate
$$ \tan 30° \cdot \tan 60°. $$Hint
Since $60° = 90° - 30°$, rewrite $\tan 60°$ using the complementary angle formula. Recall $\tan\theta \cdot \cot\theta = 1$.
Solution
Since $\tan 60° = \tan(90° - 30°) = \cot 30°$,
$$ \tan 30° \cdot \tan 60° = \tan 30° \cdot \cot 30°. $$From $\cot\theta = \dfrac{1}{\tan\theta}$,
$$ \tan 30° \cdot \cot 30° = \tan 30° \cdot \dfrac{1}{\tan 30°} = 1. $$Exercise 4
Show, using the complementary angle formula, that
$$ \sin 10° \cdot \sin 20° \cdot \sin 30° = \cos 80° \cdot \cos 70° \cdot \cos 60°. $$Hint
Apply $\sin\alpha = \cos(90° - \alpha)$ to each factor on the left.
Solution
Applying $\sin(90° - \theta) = \cos\theta$ (equivalently $\sin\alpha = \cos(90° - \alpha)$) to each factor:
$$ \sin 10° = \cos(90° - 10°) = \cos 80°, $$ $$ \sin 20° = \cos(90° - 20°) = \cos 70°, $$ $$ \sin 30° = \cos(90° - 30°) = \cos 60°. $$Multiplying these together,
$$ \sin 10° \cdot \sin 20° \cdot \sin 30° = \cos 80° \cdot \cos 70° \cdot \cos 60°, $$as required.