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,

Applying the sine addition formula

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

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.

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 \(\cos \left( {{{90}^\circ} + \beta } \right) = – \sin \beta :\)

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

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

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

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

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

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 \(\tan \left( {\large{\frac{\pi }{2}}\normalsize + \beta } \right) = – \cot \beta :\)

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

Solution.

Applying the reduction formulas yields:

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

Then

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

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:

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

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

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

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

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

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

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

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

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