Trigonometric Identity

[anemone]

Prove that $\sin^7 x=\dfrac{35\sin x-21\sin 3x+7\sin 5x-\sin 7x}{64}$.

 

[kaliprasad]

Spoiler: my Solution

We have $\sin\,x=\frac{e^{ix}-e^{-ix}}{2i}$

To avoid fraction we have

$2i\sin\,x=e^{ix}-e^{-ix}$

Take power 7

$-128i\sin^7x=(e^{ix}-e^{-ix})^7$

$={7\choose 0}e^{7ix}- {7\choose 1}e^{5ix} + {7\choose 2}e^{3ix} - {7\choose 3}e^{ix} + {7\choose 4}e^{-ix} - {7\choose 5}e^{-3ix} + {7\choose 6}e^{-5ix}- {7\choose 7}e^{-7ix}$

$={7\choose 0}e^{7ix}- {7\choose 1}e^{5ix} + {7\choose 2}e^{3ix} - {7\choose 3}e^{ix} + {7\choose 3}e^{-ix} - {7\choose 2}e^{-3ix} + {7\choose 1}e^{-5ix} - {7\choose 0}e^{-7ix}$ using ${n\choose r} = {n\choose (n-r)}$

$={7\choose 0}(e^{7ix}-e^{-7ix}) - {7\choose 1}(e^{5ix} -e^{5ix}) + {7\choose 2}(e^{3ix} - e^{-3ix}) - {7\choose 3}(e^{ix} - e^{-ix})$

or $=-64\sin ^7x = {7\choose 0}(\frac{(e^{7ix}-e^{-7ix})}{2i}) - {7\choose 1}(\frac{(e^{5ix} -e^{5ix})}{2i}) + {7\choose 2}(\frac{(e^{3ix} - e^{-3ix})}{2i}) - {7\choose 3}(\frac{(e^{ix} - e^{-ix})}{2i})$

$ =1\sin 7x-7\sin 5x+21\sin 3x-35\sin x$

Hence

$\sin^7x=\frac{-1}{64}\sin7x+ \frac{7}{64}\sin5x-\frac{21}{64}\sin3x+ \frac{35}{64}\sin\,x$