06Bcomplementary

toc = Complementary Angles =

Recall that two angles are ** complementary ** if they add to 90º. (or p /2)

math \\ \text{For example, } 60^\circ \text{ and } 30^\circ \text{ are complementary.} \\. \\ \text{In radians: } \dfrac{\pi}{3} + \dfrac{\pi}{6} = \dfrac{\pi}{2} math

Complementary Angles in Trigonometry
The __sine__ of an angle is equal to the __cosine__ of its __complement__.

math \\ . \qquad \text{In degrees, } \sin(80^\circ) = \cos(10^\circ) \\. \\ . \qquad \text{In radians: } \sin \Big(\dfrac{4\pi}{9} \Big) = \cos \Big( \dfrac{\pi}{18} \Big) math
 * For example: **

math \\ . \qquad \text{In degrees, } \sin( \theta ) = \cos(90^\circ - \theta) \\. \\ . \qquad \text{In radians: } \sin \left( \theta \right) = \cos \left( \dfrac{\pi}{2}-\theta \right) math
 * In general: **

Cotangent
The cotangent (cot) of an angle is defined as the reciprocal of the tangent.

math . \qquad \cot \left( \theta \right) = \dfrac{1}{\tan \left( \theta \right) } math

... ... The tangent of an angle is __equal__ to the cotangent of its __complement__.

math \\ . \qquad \text{In degrees, } \tan(18^\circ) = \cot(72^\circ) \\ .\\ . \qquad \text{In radians: } \tan \Big(\dfrac{\pi}{10} \Big) = \cot \Big( \dfrac{4\pi}{10} \Big) math
 * For example: **

math \\ . \qquad \text{In degrees, } \tan(\theta) = \cot(90^\circ - \theta) \\. \\ . \qquad \text{In radians: } \tan \left( \theta \right) = \cot \left( \dfrac{\pi}{2}-\theta \right) math
 * In general: **

In Other Quadrants
The same relationship holds in the other quadrants, but watch for the negatives.

In all of these examples,

math \\ . \qquad (90^\circ - \theta) + (\theta) = 90^\circ \; \; \textit{so} \\. \\ . \qquad \sin(90^\circ - \theta) = \cos(\theta) math
 * 1st Quadrant: **

math \\ . \qquad (90^\circ + \theta) + (-\theta) = 90^\circ \; \; \textit{so} \\. \\ . \qquad \sin(90^\circ + \theta) = \cos(-\theta) = \cos(\theta) math
 * 2nd Quadrant: **

math \\ . \qquad (270^\circ - \theta) + (-180^\circ + \theta) = 90^\circ \; \; \textit{so} \\. \\ . \qquad \sin(270^\circ - \theta) = \cos(-180^\circ + \theta) = -\cos(\theta) math
 * 3rd Quadrant: **

math \\ . \qquad (270^\circ + \theta) + (-180^\circ - \theta) = 90^\circ \; \; \textit{so} \\. \\ . \qquad \sin(270^\circ + \theta) = \cos(-180^\circ - \theta) = -\cos(\theta) math
 * 4th Quadrant: **


 * Notice that the calculations for ** cos ** give us the correct sign for ** sin ** in that quadrant.

We can do similar calculations for **cos** and **tan**. The results are summarised in this table:


 * Notice that adding/subtracting the angles in each entry gives __ either __ p /2 __ or __ 3 p /2.

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