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.

ParallelAngleBisector

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,

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

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

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

making use of the trigonometric periodicity identities.

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

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

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

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

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

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We have

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

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

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We have

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

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

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

$$\begin{aligned} \tan \left( \frac{\pi}{3}-A \right ) = \tan \left( \frac{\pi}{2} - \frac{\pi}{6}-A \right ). \end{aligned}$$

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

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

What is $\csc \frac{5\pi}{6} ?$

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We have

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

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

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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 $$\begin{aligned} \tan \left ( \frac{\pi}{2} - x \right ) + \cot \left ( \frac{\pi}{2} - x \right ) &= 2 \\ \cot x + \tan x &= 2 \\ \frac{1}{\tan x} + \tan x &= 2 \\ 1 + \tan^2 x &= 2\tan x \\ \tan^2 x -2\tan x +1 &= 0 \\ ( \tan x - 1)^2 &= 0 \\ \Rightarrow \tan x& = 1. \ _ \square \end{aligned}$$

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.

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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 $44\cdot 1 + \cos^2(45^\circ) = 44.5.$ $_\square$

• Trigonometry