# 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 $$\large{\frac{\pi }{2}}\normalsize$$ 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

${\frac{\pi }{2} \pm \beta ,\;\;}\kern0pt{\pi \pm \beta ,\;\;}\kern0pt{\frac{{3\pi }}{2} \pm \beta ,\;\;}\kern0pt{2\pi \pm \beta }$

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 $$\alpha = \large{\frac{{3\pi }}{2}}\normalsize,$$ 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 $$\large{\frac{{\pi }}{2}}\normalsize$$ or $$\large{\frac{{3\pi }}{2}}\normalsize,$$ 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 $$\cot \left( {\large{\frac{{3\pi }}{2}}\normalsize + \beta } \right).$$

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

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

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

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate $$\cos 840^\circ.$$

### Example 2

Calculate $$\cot \left( { – {{765}^\circ}} \right).$$

### Example 3

Find the value of the function $$\sin \large{\frac{{31\pi }}{6}}\normalsize.$$

### Example 4

Determine the value of the function $$\tan \left( { – \large{\frac{{21\pi }}{4}}\normalsize} \right).$$

### Example 5

Simplify the expression $\cos \left( {\pi – \alpha } \right)\tan \left( {\pi – \alpha } \right) – \sin \alpha .$

### Example 6

Simplify the expression ${\sin \frac{{3\pi }}{4}\cos \frac{\pi }{4} }-{ \tan \frac{{2\pi }}{3}\cos \frac{{5\pi }}{6}.}$

### Example 7

Simplify the expression ${\sin \left( {\frac{\pi }{2} – \alpha } \right)\cos \left( {\pi + \alpha } \right) }+{ \sin \left( {\pi + \alpha } \right)\sin \alpha .}$

### Example 8

Simplify the expression ${\tan \left( {2\pi – \alpha } \right)\sin \left( {2\pi – \alpha } \right) }-{ \cos \left( {\pi – \alpha } \right).}$

### Example 9

Prove that in any triangle $\sin \left( {\alpha + \beta } \right) = \sin \gamma .$

### Example 10

Prove that in any triangle $\tan \left( {\alpha + \beta } \right) = – \tan \gamma .$

### Example 1.

Calculate $$\cos 840^\circ.$$

Solution.

As $${840^\circ} = {120^\circ} + {720^\circ},$$ and cosine has a period of $$360^\circ,$$ we have

${\cos {840^\circ} = \cos \left( {{{120}^\circ} + {{720}^\circ}} \right) }={ \cos \left( {{{120}^\circ} + {{360}^\circ} \times 2} \right) }={ \cos {120^\circ} }={ \cos \left( {{{90}^\circ} + {{30}^\circ}} \right).}$

Now we use the identity $$\cos \left( {{{90}^\circ} + \beta } \right) = – \sin \beta :$$

${\cos {840^\circ} = \cos \left( {{{90}^\circ} + {{30}^\circ}} \right) }={ – \sin {30^\circ} }={ – \frac{1}{2}.}$

### Example 2.

Calculate $$\cot \left( { – {{765}^\circ}} \right).$$

Solution.

The angle $${ – {{765}^\circ}}$$ can be expressed as $$– {765^\circ} = 135^\circ – 900^\circ.$$ Therefore,

${\cot \left( { – {{765}^\circ}} \right) = \cot \left( {{{135}^\circ} – {{900}^\circ}} \right) }={ \cot \left( {{{135}^\circ} – {{180}^\circ} \times 5} \right) }={ \cot {135^\circ} }={ \cot \left( {{{90}^\circ} + {{45}^\circ}} \right).}$

Using the identity $$\cot \left( {{{90}^\circ} + \beta } \right) = – \tan \beta ,$$ we get

${\cot \left( { – {{765}^\circ}} \right) }={ \cot \left( {{{90}^\circ} + {{45}^\circ}} \right) }={ – \tan {45^\circ} }={ – 1.}$

### Example 3.

Find the value of the function $$\sin \large{\frac{{31\pi }}{6}}\normalsize.$$

Solution.

The angle $$\large{\frac{{31\pi }}{6}}\normalsize$$ can be written as

${\frac{{31\pi }}{6} = \frac{{7\pi + 24\pi }}{6} }={ \frac{{7\pi }}{6} + \frac{{24\pi }}{6} }={ \frac{{7\pi }}{6} + 4\pi .}$

Since the sine is a periodic function, with period $$2\pi,$$ we have

${\sin \frac{{31\pi }}{6} = \sin \left( {\frac{{7\pi }}{6} + 4\pi } \right) }={ \sin \left( {\frac{{7\pi }}{6} + 2\pi \cdot 2} \right) }={ \sin \frac{{7\pi }}{6} }={ \sin \left( {\pi + \frac{\pi }{6}} \right).}$

Using the reduction formula $$\sin \left( {\pi + \beta } \right) = – \sin \beta ,$$ we obtain

${\sin \frac{{31\pi }}{6} = \sin \left( {\pi + \frac{\pi }{6}} \right) }={ – \sin \frac{\pi }{6} }={ – \frac{1}{2}.}$

### Example 4.

Determine the value of the function $$\tan \left( { – \large{\frac{{21\pi }}{4}}\normalsize} \right).$$

Solution.

We represent the angle $${ – \large{\frac{{23\pi }}{4}}\normalsize}$$ in the form

${- \frac{{21\pi }}{4} = – \frac{{24\pi – 3\pi }}{4} }={ \frac{{3\pi }}{4} – \frac{{24\pi }}{4} }={ \frac{{3\pi }}{4} – 6\pi .}$

Then

${\tan \left( { – \frac{{21\pi }}{4}} \right) }={ \tan \left( {\frac{{3\pi }}{4} – 6\pi } \right) }={ \tan \frac{{3\pi }}{4} }={ \tan \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right).}$

Finally, use the cofunction identity $$\tan \left( {\large{\frac{\pi }{2}}\normalsize + \beta } \right) = – \cot \beta :$$

${\tan \left( { – \frac{{21\pi }}{4}} \right) }={ \tan \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right) }={ – \cot \frac{\pi }{4} }={ – 1.}$

### Example 5.

Simplify the expression $\cos \left( {\pi – \alpha } \right)\tan \left( {\pi – \alpha } \right) – \sin \alpha .$

Solution.

Applying the reduction formulas yields:

${\cos \left( {\pi – \alpha } \right) = – \cos \alpha ,\;\;}\kern0pt{\tan \left( {\pi – \alpha } \right) = – \tan \alpha .}$

Then

$\require{cancel}{\cos \left( {\pi – \alpha } \right)\tan \left( {\pi – \alpha } \right) – \sin \alpha }={ – \cos \alpha \cdot \left( { – \tan \alpha } \right) – \sin \alpha }={ \frac{{\cancel{\cos \alpha} \sin \alpha }}{\cancel{\cos \alpha }} – \sin \alpha }={ \cancel{\sin \alpha} – \cancel{\sin \alpha} }={ 0.}$

### Example 6.

Simplify the expression $\sin \frac{{3\pi }}{4}\cos \frac{\pi }{4} – \tan \frac{{2\pi }}{3}\cos \frac{{5\pi }}{6}.$

Solution.

Using the cofunction and reduction identities, we get

${\sin \frac{{3\pi }}{4} }={ \sin \left( {\frac{\pi }{2} + \frac{\pi }{4}} \right) }={ \cos \frac{\pi }{4} }={ \frac{{\sqrt 2 }}{2},}$

${\tan \frac{{2\pi }}{3} = \tan \frac{{4\pi }}{6} }={ \tan \frac{{3\pi + \pi }}{6} }={ \tan \left( {\frac{\pi }{2} + \frac{\pi }{6}} \right) }={ – \cot \frac{\pi }{6} }={ – \sqrt 3 ,}$

${\cos \frac{{5\pi }}{6} }={ \cos \left( {\pi – \frac{\pi }{6}} \right) }={ – \cos \frac{\pi }{6} }={ – \frac{{\sqrt 3 }}{2}.}$

Substituting these values into the original expression yields:

${\sin \frac{{3\pi }}{4}\cos \frac{\pi }{4} – \tan \frac{{2\pi }}{3}\cos \frac{{5\pi }}{6} }={ \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} – \left( { – \sqrt 3 } \right) \cdot \left( { – \frac{{\sqrt 3 }}{2}} \right) }={ \frac{1}{2} – \frac{3}{2} }={ – \frac{2}{2} }={ – 1.}$

### Example 7.

Simplify the expression ${\sin \left( {\frac{\pi }{2} – \alpha } \right)\cos \left( {\pi + \alpha } \right) }+{ \sin \left( {\pi + \alpha } \right)\sin \alpha .}$

Solution.

Since

$\sin \left( {\frac{\pi }{2} – \alpha } \right) = \cos \alpha ,$

$\cos \left( {\pi + \alpha } \right) = – \cos \alpha ,$

$\sin \left( {\pi + \alpha } \right) = – \sin \alpha ,$

we have

${\sin \left( {\frac{\pi }{2} – \alpha } \right)\cos \left( {\pi + \alpha } \right) }+{ \sin \left( {\pi + \alpha } \right)\sin \alpha }={ \cos \alpha \cdot \left( { – \cos \alpha } \right) }+{ \left( { – \sin \alpha } \right) \cdot \sin \alpha }={ – {\cos ^2}\alpha – {\sin ^2}\alpha }={ – \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) }={ – 1.}$

### Example 8.

Simplify the expression ${\tan \left( {2\pi – \alpha } \right)\sin \left( {2\pi – \alpha } \right) }-{ \cos \left( {\pi – \alpha } \right).}$

Solution.

We calculate each term separately. The tangent function is odd and has a period of $$\pi.$$ Hence,

${\tan \left( {2\pi – \alpha } \right) }={ \tan \left( { – \alpha } \right) }={ – \tan \alpha .}$

The sine function is odd and has a period 0f $$2\pi.$$ Therefore, we can write

$\sin \left( {2\pi – \alpha } \right) = \sin \left( { – \alpha } \right) = – \sin \alpha .$

The cosine function $$\cos \left( {\pi – \alpha } \right)$$ is represented as

$\cos \left( {\pi – \alpha } \right) = – \cos \alpha .$

Substituting these results, we have

${\tan \left( {2\pi – \alpha } \right)\sin \left( {2\pi – \alpha } \right) }-{ \cos \left( {\pi – \alpha } \right) }={ \left( { – \tan \alpha } \right) \cdot \left( { – \sin \alpha } \right) – \left( { – \cos \alpha } \right) }={ \frac{{\sin \alpha }}{{\cos \alpha }}\sin \alpha + \cos \alpha }={ \frac{{{{\sin }^2}\alpha }}{{\cos \alpha }} + \frac{{{{\cos }^2}\alpha }}{{\cos \alpha }} }={ \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha }} }={ \frac{1}{{\cos \alpha }} }={ \sec \alpha .}$

### Example 9.

Prove that in any triangle $\sin \left( {\alpha + \beta } \right) = \sin \gamma .$

Solution.

The sum of all angles of a triangle is equal to $$180^\circ = \pi:$$

${\alpha + \beta + \gamma = \pi ,}\;\; \Rightarrow {\alpha + \beta = \pi – \gamma .}$

Then

${\sin \left( {\alpha + \beta } \right) }={ \sin \left( {\pi – \gamma } \right).}$

By the reduction formula, we find that

${\sin \left( {\alpha + \beta } \right) }={ \sin \left( {\pi – \gamma } \right) }={ \sin \gamma .}$

### Example 10.

Prove that in any triangle $\tan \left( {\alpha + \beta } \right) = – \tan \gamma .$

Solution.

Let $$\alpha, \beta, \gamma$$ be the angles of a triangle. It is known that

$\alpha + \beta + \gamma = \pi .$

Hence,

$\alpha + \beta = \pi – \gamma .$

Using the reduction formula for tangent, we get

${\tan \left( {\alpha + \beta } \right) }={ \tan \left( {\pi – \gamma } \right) }={ – \tan \gamma .}$