Trigonometric Functions — MATHS STD 11 Science — Question

3. $\tan\frac{\alpha}{2}$

Solution:

$\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+\cos3\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha(\cos\alpha+1)}$

$=\frac{\sin\alpha}{\cos\alpha+1}$

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

$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}$

$=\frac{\sin\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}$

$=\tan\frac{\alpha}{2}$