01 / 20

Sine & Cosine of
Complementary Angles

Discover why \(\sin(\theta) = \cos(90^\circ - \theta)\) and how complementary angle identities connect trigonometric ratios in every right triangle.

Precalculus

What Are Complementary Angles?

Two angles are complementary when they add up to exactly \(90^\circ\). In every right triangle, the two acute angles are always complementary: if one is \(\theta\), the other is \(90^\circ - \theta\).

\(\theta\) \(90^\circ - \theta\) A B C adjacent to \(\theta\) opposite to \(\theta\) hypotenuse

Quick Review: Trig Ratios

For an acute angle \(\theta\) in a right triangle, the sine and cosine are defined as ratios of specific sides to the hypotenuse.

\(\sin \theta\)
\(\displaystyle\frac{\text{opposite}}{\text{hypotenuse}}\)
\(\cos \theta\)
\(\displaystyle\frac{\text{adjacent}}{\text{hypotenuse}}\)
\(\tan \theta\)
\(\displaystyle\frac{\text{opposite}}{\text{adjacent}}\)

The Big Idea: Perspective Swap

The same triangle has two acute angles. When you switch your viewpoint from one angle to the other, the roles of "opposite" and "adjacent" swap: what was opposite becomes adjacent, and vice versa.

\(\theta\) \(90^\circ-\theta\) opp to \(\theta\) = a adj to \(\theta\) = b hyp = c From \(\theta\): \(\sin \theta = \frac{a}{c}\) \(\cos \theta = \frac{b}{c}\) From \(90^\circ-\theta\): \(\sin(90^\circ-\theta) = \frac{b}{c}\) \(\cos(90^\circ-\theta) = \frac{a}{c}\)

The Cofunction Identities

These are called cofunction identities because sine and cosine are cofunctions of each other: each function of an angle equals its cofunction of the complement.

Visual Proof with a Right Triangle

\(\alpha\) \(\beta = 90^\circ-\alpha\) P Q R b a c \(\sin \alpha = \frac{a}{c}\) \(\cos \alpha = \frac{b}{c}\) \(\sin \beta = \frac{b}{c}\) \(\cos \beta = \frac{a}{c}\) \(\therefore \sin \alpha = \cos \beta\) \(\therefore \cos \alpha = \sin \beta\)

All Six Cofunction Pairs

The cofunction relationship extends to all six trigonometric functions. Each function pairs with its "co-" partner:

Function Cofunction Identity Partner
\(\sin \theta\) \(\cos(90^\circ - \theta)\) cosine
\(\cos \theta\) \(\sin(90^\circ - \theta)\) sine
\(\tan \theta\) \(\cot(90^\circ - \theta)\) cotangent
\(\cot \theta\) \(\tan(90^\circ - \theta)\) tangent
\(\sec \theta\) \(\csc(90^\circ - \theta)\) cosecant
\(\csc \theta\) \(\sec(90^\circ - \theta)\) secant

Complementary vs Supplementary

Students often confuse these two terms. The cofunction identities use complementary angles, not supplementary ones.

Complementary
Two angles summing to \(90^\circ\). This is what cofunctions use: \(\sin \theta = \cos(90^\circ - \theta)\).
Supplementary
Two angles summing to \(180^\circ\). These give a different identity: \(\sin \theta = \sin(180^\circ - \theta)\).

In Radians

Since \(90^\circ = \frac{\pi}{2}\) radians, the cofunction identities in radian form are:

The form changes but the idea stays the same: replace the angle with its complement. Whether you use degrees or radians, the complement is what remains to reach a right angle.

How to Apply Cofunction Identities

1
Identify the trig function and its angle: is it \(\sin\), \(\cos\), \(\tan\), etc.?
2
Find the complement: subtract the angle from \(90^\circ\) (or \(\frac{\pi}{2}\) in radians).
3
Switch to the cofunction: replace sine with cosine (or vice versa), tangent with cotangent, secant with cosecant.
4
Simplify if needed: evaluate the resulting expression or use it to solve an equation.

Example 1: Direct Substitution

Problem: Express \(\sin 35^\circ\) as a cosine.

1
Identify: We have \(\sin 35^\circ\). The angle is \(35^\circ\).
2
Find the complement: \(90^\circ - 35^\circ = 55^\circ\).
3
Switch to cofunction: \(\sin 35^\circ = \cos 55^\circ\).

Example 2: Solving an Equation

Problem: If \(\sin(3x + 10)^\circ = \cos(2x - 5)^\circ\), find \(x\).

1
Apply the identity: If \(\sin A = \cos B\), then \(A + B = 90^\circ\).
2
Set up the equation: \((3x + 10) + (2x - 5) = 90\).
3
Combine like terms: \(5x + 5 = 90\).
4
Solve: \(5x = 85\), so \(x = 17\).

Example 3: Triangle with Real Values

Problem: In a right triangle, the legs are \(a = 5\) and \(b = 12\), with hypotenuse \(c = 13\). Verify the cofunction identity.

\(\alpha\) \(\beta\) 12 5 13 \(\sin \alpha = \frac{5}{13}\) \(\cos \alpha = \frac{12}{13}\) \(\sin \beta = \frac{12}{13}\) \(\cos \beta = \frac{5}{13}\) \(\sin \alpha = \cos \beta\) ✓ \(\cos \alpha = \sin \beta\) ✓

Common Mistakes

✕ Using 180° instead of 90°
Writing \(\sin \theta = \cos(180^\circ - \theta)\). That is a supplementary identity, not a cofunction identity.
✕ Forgetting to switch functions
Writing \(\sin 40^\circ = \sin 50^\circ\) instead of \(\sin 40^\circ = \cos 50^\circ\). You must change sine to cosine.
✕ Subtracting from the wrong value
Computing \(\theta - 90^\circ\) instead of \(90^\circ - \theta\). For \(\theta = 25^\circ\), the complement is \(65^\circ\), not \(-65^\circ\).

Summary Table

Expression Equals (Degrees) Equals (Radians)
\(\sin 30^\circ\) \(\cos 60^\circ = \frac{1}{2}\) \(\cos\frac{\pi}{3}\)
\(\cos 30^\circ\) \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) \(\sin\frac{\pi}{3}\)
\(\sin 45^\circ\) \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) \(\cos\frac{\pi}{4}\)
\(\tan 30^\circ\) \(\cot 60^\circ = \frac{\sqrt{3}}{3}\) \(\cot\frac{\pi}{3}\)
\(\sec 20^\circ\) \(\csc 70^\circ\) \(\csc\frac{7\pi}{18}\)

Question 1 of 5

Which expression is equal to \(\sin 28^\circ\)?

Question 2 of 5

If \(\cos \theta = \sin 40^\circ\), what is the value of \(\theta\)?

Question 3 of 5

If \(\sin(2x + 15)^\circ = \cos(3x - 10)^\circ\), what is \(x\)?

Question 4 of 5

Which of the following is a correct cofunction identity?

Question 5 of 5

In a right triangle, \(\sin A = \frac{3}{5}\). What is \(\cos B\), where \(B\) is the other acute angle?