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.
References
Frequently Asked Questions
Q1: What are the complementary angle formulas?
A: The complementary angle formulas (also called cofunction identities) express trigonometric values at an angle θ in terms of values at its complement π/2 − θ (90° − θ). The three fundamental ones are: sin(π/2 − θ) = cosθ, cos(π/2 − θ) = sinθ, tan(π/2 − θ) = cotθ. The 'co-' prefix in cosine, cotangent, and cosecant comes from 'complementary'; e.g., cosine literally means 'sine of the complement'.
Q2: How do you prove the complementary angle formulas?
A: Use a right triangle. If one acute angle is θ, the other must be π/2 − θ since the angles sum to π/2. From the perspective of the second angle, what was the opposite side becomes the adjacent side and vice versa. Since sin = opposite/hypotenuse and cos = adjacent/hypotenuse, the sine of one acute angle equals the cosine of the other. The tan/cot identity follows from tan = sin/cos and substituting the sin/cos cofunction identities.
Q3: What does the 'co-' prefix in cosine mean?
A: The 'co-' in cosine, cotangent, and cosecant comes from the Latin 'complementi' (complementary). Cosine means 'complement's sine', cotangent means 'complement's tangent', and cosecant means 'complement's secant'. The complementary angle formulas are essentially the historical reason these names exist.