How To Calculate A Cofunction

Ever wonder how trigonometric functions like sine and cosine are related? They're both used for calculating sides and angles in triangles, but the relationship goes further than that. ​Cofunction identities​ give us specific formulas that show how to convert between sine and cosine, tangent and cotangent, and secant and cosecant.

An easy way to remember which functions are cofunctions is that two trig functions are ​cofunctions​ if one of them has the "co-" prefix in front of it. So:

•sine and ​co​sine are ​co​functions. 
•tangent and ​co​tangent are ​co​functions.
•secant and ​co​secant are ​co​functions.

We can calculate back and forth between cofunctions using this definition: The value of a function of an angle equals the value of the cofunction of the complement.

That sounds complicated, but instead of talking about the value of a function in general let's use a specific example. The ​sine​ of an angle equals the ​cosine​ of its complement. And the same goes for other cofunctions: The tangent of an angle equals the cotangent of its complement.

Remember: Two angles are ​complements​ if they add up to 90 degrees.

(Notice that 90° − ​x​ gives us an angle's complement.)

\(\sin(x) = \cos(90° – x) \
\cos(x) = \sin(90° – x) \
\tan(x) = \cot(90° – x) \
\cot(x) = \tan(90° – x) \
\sec(x) = \csc(90° – x)\
\csc(x) = \sec(90° – x)\)

Remember that we can also write things in terms of ​radians​, which is the SI unit for measuring angles. Ninety degrees is the same as π/2 radians, so we can also write the cofunction identities like this:

\(\sin(x) = \cos\bigg(\frac{π}{2} – x\bigg) \
\,\
\cos(x) = \sin\bigg(\frac{π}{2} – x\bigg) \
\,\
\tan(x) = \cot\bigg(\frac{π}{2} – x\bigg) \
\,\
\cot(x) = \tan\bigg(\frac{π}{2} – x\bigg) \
\,\
\sec(x) = \csc\bigg(\frac{π}{2} – x\bigg)\
\,\
\csc(x) = \sec\bigg(\frac{π}{2} – x\bigg)\)

This all sounds nice, but how can we prove that this is true? Testing it out yourself on a couple of example triangles can help you feel confident about it, but there's a more rigorous algebraic proof as well. Let's prove the cofunction identities for sine and cosine. We're going to work in radians, but it's the same as using degrees.

​Proof:​

\(\sin(x) = \cos\bigg(\frac{π}{2} – x \bigg)\)

First of all, reach way back in your memory to this formula, because we're going to use it in our proof:

\(\cos(A – B) = \cos(A)\cos(B) + \sin(A)\sin(B)\)

Got it? OK. Now let's prove: sin(​x​) = cos(π/2 − x).

We can rewrite cos(π/2 − ​x​) like this:

\(\cos\bigg(\frac{π}{2} – x\bigg) = \cos\bigg(\frac{π}{2}\bigg)\cos(x) + \sin\bigg(\frac{π}{2}\bigg)\sin(x) \
\,\)
\(\cos\bigg(\frac{π}{2} – x\bigg) = 0 × \cos(x) + 1 ×\sin(x)\)

because we know

\(\cos\bigg(\frac{π}{2}\bigg)= 0 \text{ and }\)
\(\sin\bigg(\frac{π}{2}\bigg) = 1\)

So

\(\cos\bigg(\frac{π}{2} -x\bigg) = \sin(x)\)

Ta-da! Now let's prove it with cosine!

​Proof:​

\(\cos(x)=\sin\bigg(\frac{π}{2} -x\bigg)\)

Another blast from the past: Remember this formula?

\(\sin(A – B) = \sin(A)\cos(B) – \cos(A)\sin(B)\)

We're about to use it. Now let's prove:

\(\cos(x)=\sin\bigg(\frac{π}{2} -x\bigg)\)

We can rewrite sin(π/2 − ​x​) like this:

\(\begin{aligned}
\sin\bigg(\frac{π}{2} -x\bigg) &= \sin\bigg(\frac{π}{2}\bigg)\cos(x) – \cos\bigg(\frac{π}{2}\bigg)\sin(x) \
&= 1 × \cos(x) – 0 × \sin(x)
\end{aligned}\)

because we know

\(\cos\bigg(\frac{π}{2}\bigg)= 0 \text{ and }\)
\(\sin\bigg(\frac{π}{2}\bigg) = 1\)

So we get

\(\sin\bigg(\frac{π}{2} – x\bigg) = \cos(x)\)

Cofunction Calculator

Try a few examples working with cofunctions on your own. But if you get stuck, Math Celebrity has a cofunction calculator that shows step-by-step solutions to cofunction problems.

Happy calculating!