Posted by aparajita on Saturday, January 19, 2013 at 9:05pm.
One way -- do the algebra involved in the following:
sin 5x = sin 3x cos 2x + sin 2x cos 3x
where
sin 3x = sin 2x cos x + cos 2x sin x
cos 3x = cos 2x cos x - sin 2x sin x
cos 2x = 1 - 2 sin^2 x
sin 2x = 2 sin x cos x
and cos^2 x = 1 - sin^2 x
substitute these up the line, simplify and get:
sin 5 x =16 sin^5 x - 20sin^3 x +5sin x
Another way -- combine Demoivre and the binomial theorem :
sin^5 theta is the imaginary part of ( cos theta + i sin theta)^5
(a+bi)^5 = a^5 +5a^4bi -10a^3b^2 -10a^2b^3i +5ab^4 +b^5i etc
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