# Sine & cosine of complementary angles

Learn about the relationship between the sine & cosine of complementary angles, which are angles who together sum up to 90°.

We want to prove that the sine of an angle equals the cosine of its complement.

Let's start with a right triangle. Notice how the acute angles are complementary, sum to 90$^\circ$.

Now here's the cool part. See how the sine of one acute angle

describes the $\blueD{\text{exact same ratio}}$ as the cosine of the other acute angle?

Incredible! Both functions, $\sin(\theta)$ and $\cos(90^\circ-\theta)$, give the exact same side ratio in a right triangle.

And we're done! We've shown that $\sin(\theta) = \cos(90^\circ-\theta)$.

In other words,

*the sine of an angle equals the cosine of its complement.*Well, technically we've only shown this for angles between 0$^\circ$ and 90$^\circ$. To make our proof work for all angles, we'd need to move beyond right triangle trigonometry into the world of unit circle trigonometry, but that's a task for another time.

## Cofunctions

You may have noticed that the words sine and

*co*sine sound similar. That's because they're*co*functions! The way cofunctions work is exactly what you saw above. In general, if $f$ and $g$ are cofunctions, thenand

$g(\theta) = f(90^\circ-\theta)$.

Here is a full list of the basic trigonometric cofunctions:

Cofunctions | |
---|---|

Sine and cosine | $\sin(\theta) = \cos(90^\circ-\theta)$ |

$\cos(\theta) = \sin(90^\circ-\theta)$ | |

Tangent and cotangent | $\tan(\theta) = \cot(90^\circ-\theta)$ |

$\cot(\theta) = \tan(90^\circ-\theta)$ | |

Secant and cosecant | $\sec(\theta) = \csc(90^\circ-\theta)$ |

$\csc(\theta) = \sec(90^\circ-\theta)$ |

Neat! Whoever named the trig functions must have deeply understood the relationships between them.