Trigonometric Identities

Posted on Edited on In Views:

Source: Trigonometric Identities

Pythagoras's theorem

\[ \begin{gathered} \sin ^2 \theta+\cos ^2 \theta=1 \\ 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \\ \tan ^2 \theta+1=\sec ^2 \theta \end{gathered} \]

Compound-angle formulae

\[ \begin{gathered} \cos (A+B)=\cos A \cos B-\sin A \sin B \\ \cos (A-B)=\cos A \cos B+\sin A \sin B \\ \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \sin (A-B)=\sin A \cos B-\cos A \sin B \\ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \\ \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta=2 \cos ^2 \theta-1=1-2 \sin ^2 \theta \\ \sin 2 \theta=2 \sin \theta \cos \theta \\ \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \end{gathered} \]

Sum and product formulae

\[ \begin{aligned} & \cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \\ & \cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} \\ & \sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \\ & \sin A-\sin B=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \end{aligned} \]