Trigonometry Logo

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 α and β are complementary angles, that is, two angles whose sum is 90° or π/2 radians:

\[\alpha + \beta = \frac{\pi }{2}.\]

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

Cofunction identity involving sine and cosine.
Cofunction identity cos(pi/2-beta)=sin(beta)

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:

Cofunction identity tan(pi/2-beta)=cot(beta)
Cofunction identity cot(pi/2-beta)=tan(beta)

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:

Cofunction identity sec(pi/2-beta)=csc(beta)
Cofunction identity csc(pi/2-beta)=sec(beta)

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,

Identity cos(3pi/2+beta) = sin(beta)

Applying the sine addition formula

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

yields the similar identity

Identity sin(3pi/2+beta) = -cos(beta)

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,

Reduction formula cos(pi+beta) = -cos(beta)
Reduction formula sin(pi+beta) = -sin(beta)

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,

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


Reduction formula cos(2pi-beta) = cos(beta)

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

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


Reduction formula tan(pi-beta) = -tan(beta)

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

Trigonometric cofunction and reduction identities
Figure 1.

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.


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.

Page 1 Page 2