Question

The value of $\cos A+\cos (A+B)+\dots +\cos (A+(n-1\left)B\right)=\frac{\sin \left(\frac{nB}{2}\right)\cos [A+\frac{(n-1)B}{2}]}{\sin \left(\frac{B}{2}\right)}$

• A. True
• B. False

Answer 1

The correct option is A True

$\cos A+\cos (A+B)+\cos (A+(n-1\left)B\right)+.......=C$

$\sin A+\sin (A+B)+.......\sin (A+(n-1\left)B\right)=S$

$I=\sqrt{-1}$

$C+is=e^{iA}+e^{i(A+B)}+.........e^{i(A+(n-1\left)B\right)}$

$C+is=e^{iA}[1+e^{iB}+e^{i2B}+.........e^{i\left(\right(n-1\left)B\right)}$

$C+is=e^{iA}\left[\frac{e^{inB}-1}{e^{iB}-1}\right]$ [ Geometric Prog sum of $n$ terms]

$C+is=e^{i(A+(n-1\left)\frac{B}{2}\right)}\frac{\sin \left(\frac{nB}{2}\right)}{\sin \left(B/2\right)}$ separate real terms we get

$C=\frac{\sin \frac{nB}{2}}{\sin \frac{B}{2}}\left[\cos (A+\frac{(n-1)B}{2})\right]$