# Trigonometric Co-function Identities

**Trigonometric co-function identities** are relationships between the basic trigonometric functions (sine and cosine) based on complementary angles. They also show that the graphs of sine and cosine are identical, but shifted by a constant of \(\frac{\pi}{2}\).

The identities are extremely useful when dealing with sums of trigonometric functions, as they often allow for use of the Pythagorean identities.

#### Contents

## Identities

In the diagram above, point \( P' \) is symmetric with point \(P\) with respect to the line \(y=x.\) Let \(P=(x,y)\) be the coordinates of \(P,\) then \(P'=(y,x).\)

Now, let \( \theta\) denote the angle formed by \( \overline{OP} \) and the positive direction of the \(x\)-axis. Then, since \(\overline{OP'}\) and the \(+y\)-direction also make an angle of \(\theta,\) the angle formed by \(\overline{OP'}\) and the \(+x\)-direction will be \(\frac{\pi}{2}-\theta.\) Hence the trigonometric co-functions are established as follows:

\( \sin \left( \frac{\pi}{2}-\theta \right) = \frac{x}{1} = \cos \theta \)

\( \cos \left( \frac{\pi}{2}-\theta \right) = \frac{y}{1} =\sin \theta \)

\( \tan \left( \frac{\pi}{2}-\theta \right) = \frac{x}{y}=\frac{1}{\tan \theta}=\cot \theta. \)

This is easier to see for acute angles.

Complementary angle identities:In the right triangle above,

\[ \begin{align} \cos \theta & = \frac{a}{c} = \sin \left( \frac{\pi}{2} - \theta \right) \\ \cot \theta & = \frac{a}{b} = \tan \left( \frac{\pi}{2} - \theta \right) \\ \csc \theta & = \frac{c}{b} = \sec \left( \frac{\pi}{2} - \theta \right). \end{align}\]

Additionally, since cosine is an even function and sine is an odd function,

\[\begin{align} \cos\theta &= \sin\left(\frac{\pi}{2}-\theta\right) = -\sin\left(\theta-\frac{\pi}{2}\right)=\sin\left(\theta+\frac{\pi}{2}\right)\\ \sin\theta &= \cos \left( \frac{\pi}{2}-\theta \right) = \cos \left(\theta-\frac{\pi}{2} \right)=-\cos\left(\theta+\frac{\pi}{2}\right), \end{align}\]

making use of the trigonometric periodicity identities.

Finally, the trigonometric co-functions below also directly follow:

\( \csc \left( \frac{\pi}{2}-\theta \right) = \sec \theta \)

\( \sec \left( \frac{\pi}{2}-\theta \right) = \csc \theta \)

\( \cot \left( \frac{\pi}{2}-\theta \right) =\frac{1}{\cot \theta}=\tan \theta. \)

## Example Problems

If \( \cos \left( \frac{\pi}{2}-A \right ) = \sin \frac{\pi}{4} ,\) what is \(A?\) \( \left ( 0< A < \frac{\pi}{2}\right) \)

We have

\[ \begin{align} \cos \left( \frac{\pi}{2}-A \right ) =\sin A&=\sin\frac{\pi}{4}\\ A&= \frac{\pi}{4}. \ _ \square \end{align} \]

If \( \sec \left( \frac{\pi}{2}-A \right ) = \csc \frac{\pi}{7} ,\) what is \(A?\) \( \left ( 0< A < \frac{\pi}{2}\right) \)

We have

\[ \begin{align} \sec \left( \frac{\pi}{2}-A \right )=\csc A&= \csc \frac{\pi}{7}\\ A&= \frac{\pi}{7}. \ _ \square \end{align} \]

If \( \tan \left( \frac{\pi}{3}-A \right ) = \cot \frac{\pi}{3} ,\) what is \(A?\) \( \left ( 0 < A < \frac{\pi}{2} \right ) \)

Observe that \( \tan \left( \frac{\pi}{3}-A \right ) \) can be expressed as

\[ \begin{align} \tan \left( \frac{\pi}{3}-A \right ) = \tan \left( \frac{\pi}{2} - \frac{\pi}{6}-A \right ). \end{align} \]

Then, from the trigonometric co-function identity \(\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta,\) we have

\[ \begin{align} \tan \left( \frac{\pi}{3}-A \right ) &= \tan \left( \frac{\pi}{2} - \frac{\pi}{6}-A \right ) \\ &= \tan \left( \frac{\pi}{2} -\Big( \frac{\pi}{6}+A \Big ) \right ) \\ &= \cot \left( \frac{\pi}{6} + A \right ) = \cot \left( \frac{\pi}{3} \right ) \\ \Rightarrow A &= \frac{\pi}{6}. \ _ \square \end{align} \]

What is \( \csc \frac{5\pi}{6} ?\)

We have

\[ \begin{align} \csc \frac{5\pi}{6} &= \frac{1}{\sin \frac{5 \pi}{6}} \\ &= \frac{1}{\sin \left ( \frac{\pi}{2} + \frac{\pi}{3} \right )} \\ &= \frac{1}{\sin \left ( \frac{\pi}{2} - \big (-\frac{\pi}{3} \big ) \right)} \\ &= \frac{1}{\cos \left ( -\frac{\pi}{3} \right) } \\ &= \frac{1}{\cos \frac{\pi}{3} } &\qquad \big( \text{since } \cos(-x) = \cos x \big) \\ &= 2. \ _ \square \end{align} \]

If \( \tan \left ( \frac{\pi}{2} - x \right ) + \cot \left ( \frac{\pi}{2} - x \right ) = 2,\) what is value of \( \tan x ?\)

From the trigonometric co-function identities, we know that \(\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta,\) and \(\cot\left(\frac{\pi}{2}-\theta\right)=\tan\theta.\) Hence we have \[ \begin{align} \tan \left ( \frac{\pi}{2} - x \right ) + \cot \left ( \frac{\pi}{2} - x \right ) &= 2 \\ \cot x + \tan x &= 2 \\ \frac{1}{\tan x} + \tan x &= 2 \\ 1 + \tan^2 x &= 2\tan x \\ \tan^2 x -2\tan x +1 &= 0 \\ ( \tan x - 1)^2 &= 0 \\ \Rightarrow \tan x& = 1. \ _ \square \end{align} \]

Find the value of \(\cos^2 (1^\circ) + \cos^2 (2^\circ) + \cos^2 (3^\circ) + \cdots + \cos^2 (90^ \circ) \).

Hint: Use the complementary angle identities above.

From above, we know that \(\cos\theta = \sin(90^\circ - \theta),\) so we can write \(\cos(46^\circ) = \sin(44^\circ), \cos(47^\circ) = \sin(43^\circ),\) and so on.

The sum then becomes \[\cos^2(1^\circ) + \cdots + \cos^2(44^\circ) + \cos^2(45^\circ) + \sin^2(44^\circ)+\cdots+ \sin^2(1^\circ).\] We can group the terms cleverly to obtain \[\left(\cos^2(1^\circ) + \sin^2(1^\circ)\right) + \left(\cos^2(2^\circ) + \sin^2(2^\circ)\right) + \cdots + \left(\cos^2(44^\circ) + \sin^2(44^\circ)\right) + \cos^2(45^\circ).\]

By the Pythagorean identities, \(\sin^2 x + \cos^2 x = 1,\) so our sum is simply \(44\cdot 1 + \cos^2(45^\circ) = 44.5.\) \(_\square\)

## See Also

**Cite as:**Trigonometric Co-function Identities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/trigonometric-co-function-identities/