Cofunction and Reduction Identities

Cofunction Identities

In trigonometry, a function $$f$$ is said to be a cofunction of a function $$g$$ if

$f\left( \alpha \right) = g\left( \beta \right),$

whenever $$\alpha$$ and $$\beta$$ are complementary angles, that is, two angles whose sum is $$90^\circ$$ or $$\frac{\pi }{2}$$ radians:

$\alpha + \beta = \frac{\pi }{2}.$

Using the sine and cosine subtraction formulas, we have already derived the cofunction identities

Now we will prove other similar formulas.

Consider cofunction identities involving tangent and cotangent.

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

Hence, we have the following identities:

It is also easy to deduce the cofunction identity for secant and cosecant:

$\sec \left( {\frac{\pi }{2} - \beta } \right) = \frac{1}{{\cos \left( {\frac{\pi }{2} - \beta } \right)}} = \frac{1}{{\sin \beta }} = \csc \beta ,$
$\csc \left( {\frac{\pi }{2} - \beta } \right) = \frac{1}{{\sin \left( {\frac{\pi }{2} - \beta } \right)}} = \frac{1}{{\cos \beta }} = \sec \beta .$

Thus we got two more cofunction identities:

Reduction Formulas

Actually any trigonometric function whose argument is

$\frac{\pi }{2} \pm \beta ,\;\;\pi \pm \beta ,\;\;\frac{{3\pi }}{2} \pm \beta ,\;\;2\pi \pm \beta$

can be written in terms of $$\beta.$$

Consider a few examples.

If we take the cosine addition formula

$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

and put $$\alpha = \frac{{3\pi }}{2},$$ we get

$\cos \left( {\frac{{3\pi }}{2} + \beta } \right) = \cos \frac{{3\pi }}{2}\cos \beta - \sin \frac{{3\pi }}{2}\sin \beta = 0 \cdot \cos \beta - \left( { - 1} \right) \cdot \sin \beta = \sin \beta ,$

that is,

$\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

yields the similar identity

Now let $$\alpha = \pi.$$ Using the same addition formulas we obtain the following identities:

$\cos \left( {\pi + \beta } \right) = \cos \pi \cos \beta - \sin \pi \sin \beta = \left( { - 1} \right) \cdot \cos \beta - 0 \cdot \sin \beta = - \cos \beta ,$
$\sin \left( {\pi + \beta } \right) = \sin \pi \cos \beta + \cos \pi \sin \beta = 0 \cdot \cos \beta + \left( { - 1} \right) \cdot \sin \beta = - \sin \beta ,$

that is,

Continuing in this way, we can derive all other reduction formulas.

In some cases, instead of the addition and subtraction formulas, you can use the periodicity and even/odd properties. For example, the cosine function is even and has a period of $$2\pi.$$ Therefore,

$\cos \left( {2\pi - \beta } \right) = \cos \left( { - \beta } \right) = \cos \beta ,$

or

Similarly, since tangent is odd and has a period of $$\pi,$$ we get

$\tan \left( {\pi - \beta } \right) = \tan \left( { - \beta } \right) = -\tan \beta .$

Hence,

The cofunction and reduction formulas are summarized in the table below.

The angle $$\gamma$$ denotes an original compound expression involving the angle $$\beta$$ which is supposed to be acute.

It is not necessary to memorize all these formulas! You just need to remember the following rules:

1. If the original angle $$\gamma$$ contains the angles $$\frac{{\pi }}{2}$$ or $$\frac{{3\pi }}{2},$$ the function changes to its cofunction, that is, the sine changes to cosine, tangent to cotangent, etc. If the original angle $$\gamma$$ contains $$\pi$$ or $$2\pi,$$ the function name does not change.
2. The sign of the right-hand side must correspond to the sign of trigonometric function in the left-hand side assuming that the angle $$\beta$$ is acute.

Example

Consider the function $$\cot \left( {\frac{{3\pi }}{2} + \beta } \right).$$

The angle $$\gamma = \frac{{3\pi }}{2} + \beta$$ includes $$\frac{{3\pi }}{2},$$ so cotangent changes to tangent.

Next, if $$\beta$$ is an acute angle, the angle $$\gamma = \frac{{3\pi }}{2} + \beta$$ lies in the $$4\text{th}$$ quadrant where cotangent is negative. Therefore,

$\cot \left( {\frac{{3\pi }}{2} + \beta } \right) = - \tan \beta .$

See solved problems on Page 2.