math precalc
posted by aparajita on .
find an expression for sin (5 theta) as a fifth degree polynomial in the variable sin(theta) ?

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