Question

State true or false:

$\sin \theta \cos \theta +\frac{\sin 2\theta {\cos }^2\theta }{2!}+\frac{\sin 3\theta {\cos }^2\theta }{3!}+\cdots \infty =e^{{\cos }^2\theta }\cos \left(\cos \theta \sin \theta \right)-1$. Type 1 for true and 0 for false

Answer 1

If the series has sines of multiple angles it will be taken as series S and corresponding C series we will write ourself by replacing $\sin r\theta$ by $\sin r\theta .$
$\therefore S=a\sin \theta +\frac{a^2}{2!}\sin 2\theta +\frac{a^3}{3!}\sin 3\theta +\cdots \infty$
$C=a\cos \theta +\frac{a^2}{2!}\cos 2\theta +\frac{a^3}{3!}\cos 3\theta +\cdots \infty$
where $a=\cos \theta$
$\therefore C+iS=a.e^{i\theta }+\frac{a^2}{2!}{e}^{2i\theta }+\frac{a^3}{3!}{e}^{3i\theta }+\cdots \infty$
It is of the form $x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =e^x-1$
$=e^{a{e}^{i\theta }}-1$
$=e^{a(\cos \theta +i\sin \theta )}-1$
$=e^{a\cos \theta }.e^{i\left(a\sin \theta \right)}-1$
$=e^{a\cos \theta }[\cos \left(a\sin \theta \right)+i\sin \left(a\sin \theta \right)]-1$
Equating real and imaginary parts, we get the sum of both the series.
$\therefore S=e^{a\cos \theta }\sin \left(a\sin \theta \right)$
$=e^{{\cos }^2\theta }\sin \left(\cos \theta \sin \theta \right)\because a=\cos \theta$
Also $C=e^{a\cos \theta }\cos \left(a\sin \theta \right)-1$
$=e^{{\cos }^2\theta }\cos \left(\cos \theta \sin \theta \right)-1$