Trigonometric Co-function Identities

Trigonometric co-function identitiesare relationships between thebasic trigonometric functions(sine and cosine) based oncomplementary 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 thePythagorean 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 aneven functionand sine is anodd 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 thetrigonometric 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

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

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

• Trigonometry