# Trigonometric functions of negative angles

Trigonometric functions of negative angles:

let $\theta$ and $- \theta$ be any two angles equal in magnitude but opposite in sign.

If we place each of them in the standard position it can be observed that the two angles are symmetrically placed on either side of x-axis.

Suppose we construct a circle of radius “r” with centre “o” , it will cut the terminal arms of angles $\theta$ and $- \theta$

as shown in figure below: Trigonometric functions of negative angles

Let thee points “P” and “P’” respectively. Clearly , the abscissa of P is the same as that of P’ , both in magnitude and

direction. But the ordinates of P and P’ are equal in magnitude but opposite in sign. If we denote the point

P by P(x,y) , then the point P’ will have the co-ordinates P’(x,-y), thus we have: $\sin (-\theta) = \frac{-y}{r} = - \frac{y}{r} = - \sin \theta$

And: $\cos ( - \theta ) = \frac{x}{r} = \cos \theta$

Similarly: $\tan (- \theta ) = - \tan \theta$ and $\cot (- \theta ) = - \cot \theta$

Thus: $\sin( -\theta ) = - \sin \theta , \cos ( -\theta ) = \cos \theta , \tan (- \theta ) = - \tan \theta$ and $\cot (- \theta ) = - \cot \theta$

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