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Trigonometric Function Toolkit · Tutorial 136 of 1495

Cofunction Identities from Complementary Angles

Use complementary angles to exchange sine with cosine and tangent with cotangent, and apply those relationships to exact evaluations.

Beginner 9 min read

What You'll Learn

  • Identify complementary angles in radians and degrees
  • Derive the sine-cosine cofunction identities from unit-circle coordinates
  • State the domain conditions for the tangent-cotangent identities
  • Use cofunction identities to evaluate familiar angles exactly
  • Recognize a common mistaken use of complementary-angle identities

Introduction

Some trigonometric expressions involve angles whose measures add to a right angle. Such angles are called complementary. Their trigonometric values are closely related: the sine of one angle equals the cosine of its complement, and tangent and cotangent are paired in the same way.

These relationships are called cofunction identities. They follow naturally from the unit circle: exchanging the two coordinates of a point corresponds to reflecting it across the line \(y=x\), which changes its angle to its complement. We will derive the identities from this geometric picture, state the conditions under which the quotient identities are defined, and use them to evaluate exact values.

Complementary Angles

Two angles are complementary when their measures add to a right angle. In radians, a right angle has measure \(\frac{\pi}{2}\); in degrees, it has measure \(90^\circ\). Thus the complement of an angle \(\theta\) is \(\frac{\pi}{2}-\theta\) in radians, or \(90^\circ-\theta\) in degrees.

Definition: Angles \(\theta\) and \(\frac{\pi}{2}-\theta\) are complementary because their sum is \(\theta+\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}\). In degree measure, the corresponding pair is \(\theta\) and \(90^\circ-\theta\).

The identities below are not restricted to acute angles. The same relationships hold for any real \(\theta\), with the usual care that tangent and cotangent are defined only when their denominators are nonzero. When an angle is outside the first quadrant, its sine or cosine may be negative; the cofunction identities still retain the same signs because they express equality for the actual angles, not just equality of lengths.

Why Sine and Cosine Exchange

On the unit circle, the point corresponding to an angle \(\theta\) has coordinates \((\cos\theta,\sin\theta)\). Reflecting this point across the line \(y=x\) exchanges its coordinates, giving \((\sin\theta,\cos\theta)\).

The line \(y=x\) makes an angle of \(\frac{\pi}{4}\) with the positive horizontal axis. Reflection across a line at angle \(\frac{\pi}{4}\) sends a direction at angle \(\theta\) to a direction at angle \(2\left(\frac{\pi}{4}\right)-\theta=\frac{\pi}{2}-\theta\). Therefore the reflected point is also the unit-circle point for the complementary angle \(\frac{\pi}{2}-\theta\). Since the coordinates of a unit-circle point at angle \(\frac{\pi}{2}-\theta\) are \(\left(\cos\left(\frac{\pi}{2}-\theta\right),\sin\left(\frac{\pi}{2}-\theta\right)\right)\), matching coordinates gives the two identities.

Theorem: For every real angle \(\theta\), $$ \sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta \qquad\text{and}\qquad \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta. $$ In degree measure, replace \(\frac{\pi}{2}\) by \(90^\circ\).

This argument is a complete geometric derivation: reflection exchanges the coordinates, and its effect on the angle is to replace \(\theta\) with \(\frac{\pi}{2}-\theta\). The first coordinate equality says that the cosine of the complement equals the original sine; the second says that the sine of the complement equals the original cosine.

Deriving the Tangent and Cotangent Identities

Tangent and cotangent are quotients of the unit-circle coordinates. Where their denominators are nonzero, \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\cot\theta=\frac{\cos\theta}{\sin\theta}\). The sine-cosine identities therefore produce the tangent-cotangent identities.

$$ \tan\left(\frac{\pi}{2}-\theta\right) = \frac{\sin\left(\frac{\pi}{2}-\theta\right)} {\cos\left(\frac{\pi}{2}-\theta\right)} = \frac{\cos\theta}{\sin\theta} = \cot\theta. $$

This calculation is valid when \(\sin\theta\ne0\). That is precisely the condition that both sides are defined: \(\cot\theta\) requires a nonzero sine, and the denominator of the complementary tangent is \(\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta\).

Likewise, when \(\cos\theta\ne0\),

$$ \cot\left(\frac{\pi}{2}-\theta\right) = \frac{\cos\left(\frac{\pi}{2}-\theta\right)} {\sin\left(\frac{\pi}{2}-\theta\right)} = \frac{\sin\theta}{\cos\theta} = \tan\theta. $$

Here \(\cos\theta\ne0\) is required on both sides: it is the denominator in \(\tan\theta\), and it equals \(\sin\left(\frac{\pi}{2}-\theta\right)\), the denominator in the complementary cotangent.

Theorem: The quotient cofunction identities are $$ \tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta \quad\text{when }\sin\theta\ne0, \qquad \cot\left(\frac{\pi}{2}-\theta\right)=\tan\theta \quad\text{when }\cos\theta\ne0. $$ The stated conditions matter: the identities do not assign values where either side is undefined.

The complete set of cofunction identities and their domain requirements is summarized below.

Function of the complementEqual functionCondition
\(\sin\left(\frac{\pi}{2}-\theta\right)\)\(\cos\theta\)All real \(\theta\)
\(\cos\left(\frac{\pi}{2}-\theta\right)\)\(\sin\theta\)All real \(\theta\)
\(\tan\left(\frac{\pi}{2}-\theta\right)\)\(\cot\theta\)\(\sin\theta\ne0\)
\(\cot\left(\frac{\pi}{2}-\theta\right)\)\(\tan\theta\)\(\cos\theta\ne0\)

Worked Exact Evaluations

Worked Example: Evaluate \(\cos\left(\frac{\pi}{3}\right)\) using a complement

The complement of \(\frac{\pi}{3}\) is \(\frac{\pi}{2}-\frac{\pi}{3}\). With common denominator \(6\), this is \(\frac{3\pi}{6}-\frac{2\pi}{6}=\frac{\pi}{6}\). The identity \(\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta\), used with \(\theta=\frac{\pi}{6}\), gives

$$ \cos\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}. $$

Check: the unit-circle point at angle \(\frac{\pi}{3}\) has horizontal coordinate \(\frac{1}{2}\), so its cosine is \(\frac{1}{2}\), as obtained. The familiar exact value \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\) follows, for example, from a 30-60-90 triangle scaled to have hypotenuse \(1\).

Worked Example: Evaluate \(\tan\left(\frac{\pi}{3}\right)\) as a cotangent

The complement of \(\frac{\pi}{3}\) is \(\frac{\pi}{6}\). The identity \(\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta\), with \(\theta=\frac{\pi}{6}\), gives

$$ \tan\left(\frac{\pi}{3}\right) = \cot\left(\frac{\pi}{6}\right) = \frac{\cos\left(\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{6}\right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2}\cdot 2 = \sqrt{3}. $$

The condition holds because \(\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}\ne0\). Check directly using the unit-circle coordinates at \(\frac{\pi}{3}\): the vertical coordinate is \(\frac{\sqrt{3}}{2}\) and the horizontal coordinate is \(\frac{1}{2}\), so

$$ \tan\left(\frac{\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}. $$

The direct quotient agrees with the cofunction calculation.

Worked Example: Evaluate an expression involving two complementary angles

Evaluate \(\sin\left(\frac{\pi}{2}-\frac{\pi}{3}\right)+\cot\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\). Apply a cofunction identity to each term:

$$ \sin\left(\frac{\pi}{2}-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, $$
$$ \cot\left(\frac{\pi}{2}-\frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}. $$

Thus the sum is

$$ \frac{1}{2}+\frac{\sqrt{3}}{3} = \frac{3}{6}+\frac{2\sqrt{3}}{6} = \frac{3+2\sqrt{3}}{6}. $$

Check by evaluating the original complementary angles directly: \(\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}\), and \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\). The original expression is therefore \(\sin\left(\frac{\pi}{6}\right)+\cot\left(\frac{\pi}{3}\right)=\frac{1}{2}+\frac{1}{\sqrt{3}}=\frac{3+2\sqrt{3}}{6}\), which matches.

A Common Mistake

A complement changes the angle, and the corresponding function changes too. It is incorrect to assume that changing an angle to its complement leaves the same trigonometric function unchanged. For example, the following calculation is wrong:

$$ \sin\left(\frac{\pi}{2}-\frac{\pi}{6}\right) \overset{\text{wrong}}{=} \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}. $$

The angle on the left is \(\frac{\pi}{3}\), not \(\frac{\pi}{6}\). The sine-cosine identity gives the correct calculation:

$$ \sin\left(\frac{\pi}{2}-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}. $$

For a direct check, \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), not \(\frac{1}{2}\). A useful reminder is that complementary angles exchange sine and cosine, and exchange tangent and cotangent; they do not generally preserve the function name.

Also check domains before using a tangent-cotangent identity. For \(\theta=0\), the expression \(\tan\left(\frac{\pi}{2}-\theta\right)\) is \(\tan\left(\frac{\pi}{2}\right)\), which is undefined, and \(\cot\theta=\cot 0\) is also undefined. The identity relates the two functions where they exist; it does not make either one defined at excluded angles.

Why Cofunction Identities Are Useful

Cofunction identities allow an expression involving one trigonometric function to be rewritten using its paired function. This is especially useful when an angle is naturally written as a complement, or when a familiar exact value is available for the complementary angle. For instance, evaluating \(\cos\left(\frac{\pi}{3}\right)\) can be reduced to the standard value of \(\sin\left(\frac{\pi}{6}\right)\).

They also provide a way to check whether an expression has been interpreted correctly. If the angles add to \(\frac{\pi}{2}\), sine and cosine should exchange; tangent and cotangent should exchange when defined. If a calculation keeps the same function without another justification, it is worth checking the angle and the identity again.

Key takeaway: Complementary angles add to \(\frac{\pi}{2}\) (or \(90^\circ\)). On the unit circle, reflection across \(y=x\) exchanges coordinates, giving the sine-cosine identities; taking coordinate quotients gives the tangent-cotangent identities wherever the denominators are nonzero.

Check Your Understanding

Use the cofunction identities and exact unit-circle values where needed.

  1. Write \(\cos\left(\frac{\pi}{2}-\theta\right)\) as a function of \(\theta\), and state whether a domain restriction is needed.
  2. For what condition on \(\theta\) is \(\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta\) defined?
  3. Evaluate \(\sin\left(\frac{\pi}{3}\right)\) by using its complementary angle.
  4. Evaluate \(\cot\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\) exactly.
  5. Explain why \(\sin\left(\frac{\pi}{2}-\theta\right)=\sin\theta\) is not a cofunction identity, and give an angle at which the two sides differ.

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