Multiple Angle Formulas

The trigonometric functions of multiple angles is the multiple angle formula. Double and triple angles formula are there under the multiple angle formulas. Sine, tangent and cosine are the general functions for the multiple angle formula.

The sin formula for multiple angle is:

$\large sin \theta = \sum_{k=0}^{n}\;cos^{k}\theta \; Sin^{n-k}\theta\; Sin\left [\frac{1}{2}\left(n-k\right)\right]\pi$

Where n=1,2,3,……

General formulas are,

$\large sin^{2}\theta =2 \times cos\,\theta \; sin\,\theta$

$\large sin^{3}\theta =3 \times cos^{2}\,\theta \; sin\, \theta \; sin^{3}\,\theta$

The multiple angle’s Cosine formula is given below:

$\large Cos\;n\, \theta =\sum_{k=0}^{n}cos^{k}\theta \,sin^{n-k}\theta \;cos\left [\frac{1}{2}\left(n-k\right)\pi\right]$

Where n = 1,2,3

The general formula goes as:

$\large cos^{2}\, \theta =cos^{2}\, \theta – sin^{2}\, \theta$

$\large cos^{3}\, \theta =cos^{3}\, \theta – cos\, \theta \; sin^{2}\, \theta$

Tangent Multiple Angles formula

$\large Tan\;n\theta = \frac{sin\;n\theta}{cos\;n\theta}$

Solved Examples

Question 1: Prove that $\frac{sin\,x+sin\,2x}{1+cos\,x+cos\,2x}=tan\,x$

Solution:

Using the identities and formulas above we can solve the question as follows:

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

$=\frac{sin\,x+2\,sin\,x\;cos\,x}{2+cos^{2}\,x+cos\,x}$

$=\frac{sin\,x(1+2cos\,x)}{cos\,x(2\,cos\,+1)}=tan\;x$