Question

If $S_n=(\frac{n}{1})\sin a+(\frac{n}{2})\sin 2a+\cdots +(\frac{n}{n})\sin na$ and $T_n=(\frac{n}{1})\cos a+(\frac{n}{2})\cos 2a+\cdots +(\frac{n}{n})\cos na$
where $n\in N$ and $a$ be the non-zero real number such that $a\ne (2n-1)\frac{\pi }{2}$, then which of the following is/are CORRECT?
$(\text{Here},(\frac{n}{r})={}^n{C}_r)$

• A. $S_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\sin \left(20a\right)$
• B. $S_{40}=2^{39}{\cos }^{40}\left(\frac{a}{2}\right)\sin \left(20a\right)$
• C. $T_{40}=2^{39}{\cos }^{40}\left(\frac{a}{2}\right)\cos \left(20a\right)-1$
• D. $T_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\cos \left(20a\right)-1$

Answer 1

Answer: (A), (D)

The correct options are
A $S_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\sin \left(20a\right)$
D $T_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\cos \left(20a\right)-1$

Let $z$ be a complex number such that $z=\cos a+i\sin a$
Then, by De Moivre's theorem, we have
$z^n=\cos na+i\sin na$

Now,
$1+T_n+i{S}_n=1+(\frac{n}{1})(\cos a+i\sin a)+(\frac{n}{2})(\cos 2a+i\sin 2a)+\cdots +(\frac{n}{n})(\cos na+i\sin na)$
$=(\frac{n}{0})z^0+(\frac{n}{1})z+(\frac{n}{2})z^2+...+(\frac{n}{n})z^n$
$\Rightarrow 1+T_n+i{S}_n=(1+z{)}^n\cdots (1)$

We know that,
$1+z=1+\cos a+i\sin a=2{\cos }^2\frac{a}{2}+2i\sin \frac{a}{2}\cos \frac{a}{2}=2\cos \frac{a}{2}(\cos \frac{a}{2}+i\sin \frac{a}{2})$
$\therefore (1+z{)}^n=2^n{\cos }^n\frac{a}{2}(\cos \frac{na}{2}+i\sin \frac{na}{2})\cdots (2)$

Using equation $\left(1\right)$ and $\left(2\right)$,
$(1+T_n)+i{S}_n=\left(2^n{\cos }^n\frac{a}{2}\cos \frac{na}{2}\right)+i\left(2^n{\cos }^n\frac{a}{2}\sin \frac{na}{2}\right)$
$\Rightarrow S_n=2^n{\cos }^n\frac{a}{2}\sin \frac{na}{2}$
and $T_n=2^n{\cos }^n\frac{a}{2}\cos \frac{na}{2}-1$

$\therefore S_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\sin \left(20a\right)$
and $T_{40}=2^{40}{\cos }^{40}\left(\frac{a}{2}\right)\cos \left(20a\right)-1$