Question

If $c_r$ stand by ${}^n{C}_r$ prove that
$C_{1}sin\alpha cos(n-1)\alpha +C_{2}sin2\alpha cos(n-2)\alpha +C_{3}sin3\alpha cos(n-3)\alpha +..+C_{n}sinn\alpha$

$=2^{n-1}sinn\alpha$

Answer 1

Let the given series be denoted by
E = S +$C^{n}sinn\alpha$
Where $S=C_{1}sin\alpha cos(n-1)\alpha +c_{2}sin2\alpha cos(n-2)\alpha =..$
Writing the above in reverse order , we get
S = $C_{n}cos\alpha sin(n-1)\alpha +C_{n-1}cos2\alpha sin(n-2)\alpha$
Adding 2S = $(C_1+C_n)sin(n\alpha -\alpha +\alpha )+(C_2+C_{n-1})sin(n\alpha -2\alpha +2\alpha )$
or 2S = $(C_1+C_2+C_3+...+C_{n-1})sinn\alpha$
$\therefore sinAcosB+cosAsinB=sin(A+B)+sinn\alpha$
Add and subtract $(C_0÷C_1)sinn\alpha$
$\therefore 2S=\left({\sum }_{i=0}^n{c}_i\right)sinn\alpha -(1+1)sinn\alpha$
$=(2^n-2)sinnl\alpha$
$\therefore S=(2^{2n-1}-1)sinn\alpha$
$\therefore E=s+{}^n{C}_{n}sinn\alpha$
$=S+sinn\alpha =26n-1sinn\alpha$ by (2)