Bashing Unavailable Bcoz Of Summation!

Set x=0 . This gives a0+a1+a2+a3+a4+a5+a6 = sin(0)^6=0

To express $cos^{n}x$ or $sin^{n}x$ as a series of multiple angles of cos and sin , we use $Eulers -representation$

$cos(x)$ = $\dfrac{e^{ix}+e^{-ix}}{2}$ and $sin(x)$ = $\dfrac{e^{ix}-e^{-ix}}{2i}$

And Expand $cos^{n}x$ =( $\dfrac{e^{ix}+e^{-ix}}{2})^{n}$ By Binomial theorem etc.as for the above example

$sin^{6}x$ =( $\dfrac{e^{ix}-e^{-ix}}{2i})^{6}$

$\dfrac{1}{2^6.i^6}$ [ $6C_{0}e^{6ix}-6C_{1}e^{5ix}e^{-ix}+6C_{2}e^{4ix}e^{-2ix}.........................$ ]

$=$ $\dfrac{-1}{64}$ [ $2cos(6x)-6*2cos(4x)+15*2cos(2x)-20$ ]

$\dfrac{-1}{32}$ + $\dfrac{3}{16}cos4x$ - $\dfrac{15}{32}cos2x$ + $\dfrac{5}{8}$

= $\sum_{r=0}^6a_{r}.cosrx$

$a_{0}$ =5/8

$a_{1}$ =0

$a_{2}$ =-15/32

$a_{3}$ =0

$a_{4}$ =3/16

$a_{5}$ =0

$a_{6}$ =-1/32

sum of all these =

$\dfrac{5}{16}$

$m+n$ = $\boxed{21}$

Putting x=0 we get , a1+a2+a3+a4+a5+a6 = 0