TRIGONOMETRIC RATIOS OF ALLIED ANGLES

Two angles are allied if their sum or difference is a multiple of 90o.

If \theta is any angle, then - \theta ,90^\circ  \pm \theta ,180^\circ  \pm \theta ,270^\circ  \pm \theta ,360^\circ  \pm \theta etc are called as allied angles.

General trigonometric ratios and allied terms are listed below:

\sin \left( { - \theta } \right) =  - \sin \theta

\cos \left( { - \theta } \right) = \cos \theta

\sin \left( {90^\circ  - \theta } \right) = \cos \theta

\cos \left( {90^\circ  - \theta } \right) = \sin \theta

\sin \left( {90^\circ  + \theta } \right) = \cos \theta

\cos \left( {90^\circ  + \theta } \right) = \sin \theta

\sin \left( {180^\circ  - \theta } \right) = \sin \theta

\cos \left( {180^\circ  - \theta } \right) =  - \cos \theta

\sin \left( {180^\circ  + \theta } \right) =  - \sin \theta

\cos \left( {180^\circ  + \theta } \right) =  - \cos \theta

\sin \left( {270^\circ  - \theta } \right) =  - \cos \theta

\cos \left( {270^\circ  - \theta } \right) =  - \sin \theta

\sin \left( {270^\circ  + \theta } \right) =  - \cos \theta

\cos \left( {270^\circ  + \theta } \right) = \sin \theta

\tan \left( {90^\circ  - \theta } \right) = \cot \theta

\cot \left( {90^\circ  - \theta } \right) = \tan \theta

Example:

The value of trigonometric function \cos \left( { - 45^\circ {\rm{ }}} \right) is;

We know that, \cos \left( { - \theta } \right) = \cos \theta , then;

\cos \left( { - 45^\circ {\rm{ }}} \right) = \cos 45^\circ

Question 1:

Find the value of trigonometric function \cot \left( { - 50^\circ {\rm{ }}} \right)

Solution

We know that, \cot \left( { - \theta } \right) =  - \cot \theta , then;

\cot \left( { - 50^\circ {\rm{ }}} \right) =  - \cot 50^\circ

Question 2:

Find the value of trigonometric function\cos \left( {230^\circ {\rm{ }}} \right)

Solution

\cos \left( {230^\circ {\rm{ }}} \right) is written as\cos \left( {180^\circ  + 50^\circ {\rm{ }}} \right)

Known that, \cos \left( {180^\circ  + \theta } \right) =  - \cos \theta , then;

\cos \left( {180^\circ  + 50^\circ {\rm{ }}} \right) =  - \cos 50^\circ

Post Author: E-Maths