Question

If $\frac{-\pi }{2}<\theta <\frac{\pi }{2}$, then $1+\frac{1}{2}\cos 2\theta -\frac{1}{2.4}\cos 4\theta +\frac{1.3}{\mathrm{2.4.6}}\cos 6\theta .....\infty$ equals

• A. $\sqrt{\cos \theta (1+\cos \theta )}$
• B. $\sqrt{\cos \theta \cos \frac{\theta }{2})}$
• C. $\sqrt{2\cos \theta \cos \frac{\theta }{2})}$
• D. None of these

Answer 1

The correct option is A $\sqrt{\cos \theta (1+\cos \theta )}$

Let $C=1+\frac{1}{2}\cos 2\theta -\frac{1}{2.4}\cos 4\theta +\frac{1.3}{\mathrm{2.4.6}}\cos 6\theta -.......\propto$ and

$S=1+\frac{1}{2}\sin 2\theta -\frac{1}{2.4}\sin 4\theta +\frac{1.3}{\mathrm{2.4.6}}\sin 6\theta -.......\propto$

$\therefore C+iS=1+\frac{1}{2}{e}^{i2\theta }-\frac{1}{2.4}{e}^{i4\theta }+\frac{1.3}{\mathrm{2.4.6}}\sin 6\theta .......\propto$

$=(1+e^{i2\theta }{)}^{\frac{1}{2}}=(1+\cos 2\theta +i\sin 2\theta {)}^{\frac{1}{2}}=(2{\cos }^2\theta +2i\sin \theta \cos \theta {)}^{\frac{1}{2}}$

$=\sqrt{2\cos \theta }(\cos \theta +i\sin \theta {)}^{\frac{1}{2}}(\because \frac{-\pi }{2}<0<\frac{\pi }{2})$

$=\sqrt{2\cos \theta }(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})$

Equating real part $=\sqrt{2\cos \theta \frac{(1+\cos \theta )}{2}}\therefore C=\sqrt{\cos \theta (1+\cos \theta )}$