Question

If $(\sin {\theta }_1+i\cos {\theta }_1)(\sin {\theta }_2+i\cos {\theta }_2)...(\sin {\theta }_n+i\cos {\theta }_n)=a+ib$, then $a^2+b^2=$(Given $a$ and $b$ are real numbers)

• A. $4$
• B. $2$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

We know that
$sin\theta +icos\theta$
$=e^{i(\frac{\pi }{2}-\theta )}$
$=e^{i\frac{\pi }{2}}.e^{-i\theta }$...(i)
Hence
$(sin{\theta }_1+icos{\theta }_1)(sin{\theta }_2+icos{\theta }_2)(sin{\theta }_3+icos{\theta }_3)...(sin{\theta }_n+icos{\theta }_n)$
$=e^{i(\frac{\pi }{2}-{\theta }_1)}.e^{i(\frac{\pi }{2}-{\theta }_2)}.e^{i(\frac{\pi }{2}-{\theta }_3)}...e^{i(\frac{\pi }{2}-{\theta }_n)}$
$=e^{i\left(n\right(\frac{\pi }{2})-({\theta }_1+{\theta }_2+...{\theta }_n\left)\right)}$
$=e^{i\left(\right(\frac{n\pi }{2})-({\theta }_1+{\theta }_2+...{\theta }_n\left)\right)}$
$=cos\left(\right(\frac{n\pi }{2})-({\theta }_1+{\theta }_2+...{\theta }_n\left)\right)+isin\left(\right(\frac{n\pi }{2})-({\theta }_1+{\theta }_2+...{\theta }_n\left)\right)$
$=cos(\frac{n\pi }{2}-\sum {\theta }_i)+isin(\frac{n\pi }{2}-\sum {\theta }_i))$
Hence
$a+ib=cos(\frac{n\pi }{2}-\sum {\theta }_i)+isin(\frac{n\pi }{2}-\sum {\theta }_i))$
$a=cos(\frac{n\pi }{2}-\sum {\theta }_i)$ and
$b=sin(\frac{n\pi }{2}-\sum {\theta }_i)$
Hence
$a^2+b^2=1$