Double Angle Formulas

Double-angle formulas can be expanded to multiple-angle functions (triple, quadruple, quintuple, and so on) by using the angle sum formulas, and then reapplying the double-angle formulas.

$\large sin(A+B)=sinA\;cosB+cosA\;sinB$

$\large sin(A-B)=sinA\;cosB-cosA\;sinB$

$\large cos(A+B)=cosA\;cosB-sinA\;sinB$

$\large cos(A-B)=cosA\;cosB+sinA\;sinB$

$\large sin\alpha +sin\beta =2sin\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}$

$\large sin\alpha -sin\beta =2sin\frac{\alpha -\beta }{2}cos\frac{\alpha +\beta }{2}$

$\large cos\alpha +cos\beta =2cos\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}$

$\large cos\alpha -cos\beta =-2sin\frac{\alpha +\beta }{2}sin\frac{\alpha -\beta }{2}$

$\large sin2\alpha =2\;sin\alpha\;cos\alpha$

$\large cos2\alpha =cos^{2}\alpha -sin^{2}\alpha = 2cos^{2}\alpha -1=1-2sin^{2}\alpha$

$\large tan2\alpha =\frac{2tan\alpha }{1-tan^{2}\alpha }$

Half Angle Formulas

$\large sin\left ( \frac{a}{2} \right )=\pm \sqrt{\frac{(1-cos\;a)}{2}}$

$\large cos\left ( \frac{a}{2} \right )=\pm \sqrt{\frac{(1+cos\;a)}{2}}$

$\large tan\left ( \frac{a}{2} \right )=\frac{1-cos\;a}{sin\;a}=\frac{sin\;a}{1+cos\;a}$

$\large sin\left ( \frac{a}{2} \right )=\sqrt{\frac{1-cos(a)}{2}}$

$\large cos\left ( \frac{a}{2} \right )=\sqrt{\frac{1+cos(a)}{2}}$

$\large tan\left ( \frac{a}{2} \right )=\sqrt{\frac{1-cos(a)}{1+cos(a)}}$

Practise This Question

A thin wire of length L and uniform linear mass density ρ Is bent into a circular loop with centre at 0 as shown in the figure. The moment of inertia of the loop about the axis XX’ is