Trigonometric Identities

1. Single-Angle

Interrelations

$\tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{1}{\tan \theta },\sec \theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta }$

Cofunction formulas

$\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta ,\cos \left(\frac{\pi }{2}-\theta \right)=\sin \theta ,\cot \left(\frac{\pi }{2}-\theta \right)=\tan \theta$

Pythagorean Identities

${\sin }^2\theta +{\cos }^2\theta =1,{\tan }^2\theta +1={\sec }^2\theta ,{\cot }^2\theta +1={\csc }^2\theta$

Eg.

$\frac{\sin \theta }{1-\cos \theta }=\frac{\sin \theta }{(1-\cos \theta )}\cdot \frac{(1+\cos \theta )}{(1+\cos \theta )}$

$=\frac{\sin \theta (1+\cos \theta )}{1-{\cos }^2\theta }=\frac{\sin \theta (1+\cos \theta )}{{\sin }^2\theta }=\frac{\sin \theta (1+\cos \theta )}{\sin \theta \cdot \sin \theta }=\frac{1+\cos \theta }{\sin \theta }$

Prove the following identities

a) ${\cos }^2\theta {\tan }^3\theta =\tan \theta -\sin \theta \cos \theta$

${\cos }^2\theta {\tan }^3\theta ={\cos }^2\theta \cdot \frac{{\sin }^3\theta }{{\cos }^3\theta }=\frac{{\sin }^3\theta }{\cos \theta }=\frac{\sin \theta \cdot {\sin }^2\theta }{\cos \theta }$

$=\tan \theta {\sin }^2\theta =\tan \theta (1-{\cos }^2\theta )=\tan \theta -\tan \theta {\cos }^2\theta$

$=\tan \theta -\frac{\sin \theta }{\cos \theta }{\cos }^2\theta =\tan \theta -\sin \theta \cos \theta$

b) $\frac{1}{\sec \theta -\tan \theta }=\sec \theta +\tan \theta$

$\frac{1}{\sec \theta -\tan \theta }=\frac{1}{\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}=\frac{1}{\frac{1-\sin \theta }{\cos \theta }}=\frac{\cos \theta }{1-\sin \theta }\cdot \frac{(1+\sin \theta )}{(1+\sin \theta )}$

$=\frac{\cos \theta +\sin \theta \cos \theta }{1-{\sin }^2\theta }=\frac{\cos \theta +\sin \theta \cos \theta }{{\cos }^2\theta }$

$=\frac{\cos \theta }{{\cos }^2\theta }+\frac{\sin \theta \cos \theta }{{\cos }^2\theta }=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\sec \theta +\tan \theta$

c) $\csc \theta -\cot \theta =\frac{\sin \theta }{1+\cos \theta }$

$\csc \theta -\cot \theta =\frac{1}{\sin \theta }-\frac{1}{\tan \theta }=\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }=\frac{1-\cos \theta }{\sin \theta }$

$=\frac{(1-\cos \theta )}{\sin \theta }\cdot \frac{(1+\cos \theta )}{(1+\cos \theta )}=\frac{1-{\cos }^2\theta }{\sin \theta (1+\cos \theta )}=\frac{{\sin }^2\theta }{\sin \theta (1+\cos \theta )}=\frac{\sin \theta }{1+\cos \theta }$

Double Angle Formulas

$\sin 2\theta =2\sin \theta \cos \theta$

$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =2{\cos }^2\theta -1=1-2{\sin }^2\theta$

Half Angle Formulas

${\sin }^2\frac{\theta }{2}=\frac{1-\cos \theta }{2},{\cos }^2\frac{\theta }{2}=\frac{1+\cos \theta }{2}$

$\tan \frac{\theta }{2}=\frac{1-\cos \theta }{1+\cos \theta },\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }=\frac{1-\cos \theta }{\sin \theta }$

Double-Angle Identities

$\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

$\cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$

$\tan (\alpha \pm \beta )=\frac{\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }$

Geometric Relationships

The construction at the bottom of the page proves the compound-angle formulas geometrically, with the labelled lengths: $\cos \alpha \cos \beta$ along the base, $\cos \beta$ on the inner hypotenuse, $\sin \beta$ and $\cos \alpha \sin \beta$ on the right, $\sin \alpha \sin \beta$, and the outer triangle’s sides $\sin (\alpha +\beta )$ and $\cos (\alpha +\beta )$ subtending the angle $\alpha +\beta$.