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Proving Using Mathematical Induction

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This is the Lagrange's trig identity. I think the key to prove by induction is to convert sin (n+3/2)x into sin ((n+1) + 1/2) x = sin (n+1)x cos (x/2) + cos (n+1)x * sin (x/2), but to also do the same for sin (n+1/2)x as sin ((n+1) - 1/2)x
Proof f(n)--> f(n+1)

 sin (n+1/2)x / (2 sin (x/2)) + cos (n+1)x ... use sin (a - b), where a = (n + 1)x, b = x/2

= [sin (n+1)x cos (x/2) - cos (n+1)x sin (x/2) + 2 cos(n+1)x sin (x/2)] / (2 sin (x/2)) 

= [sin (n+1)x cos(x/2) + cos(n+1)x sin (x/2)] / (2 sin (x/2))
= sin (n+3/2) x / (2 sin (x/2))

If you are trying to prove f(1), then maybe tan (x / 2) = sin x / (1 + cos x) is helpful.

Edited by kw0573

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