[AxS.pinealan]

Hey I got a trigonometry question for you guys:
Use mathematical induction to prove that for any positive integer n,
sinx+sin3x+...+sin(2n-1)x= (1-cos2nx)/2sinx
where sinx =/= 0

[Mukade]

Originally Posted by pinealan

Hey I got a trigonometry question for you guys:
Use mathematical induction to prove that for any positive integer n,
sin x+sin 3x+...+sin(2n-1)x= (1-cos2nx)/2sin x
where sin x =/= 0

I'll use the indentity (1) (sin x)^2= (1-cos 2x)/2
because (sin nx)^2/(sin x) is easier to work with.
and also use
(2) (sin A sin B=1/2(cos(A-B)-cos(A+B))


Set n=1

lhs=sinx
rhs=(sin x)^2/sin x
=sin x
=lhs
true for n=1

Assume n=k
sin x+ sin 3x+...+sin(2k-1)x=(sin kx)^2/(sin x)

For n=k+1

lhs= ((sin kx)^2)/sin x+sin(2k+1)x
=(((sin kx)^2)+sin(2k+1)x sin x)/sin x
=(((sin kx)^2)+1/2(cos(x-2kx-x)-cos(2kx+2x))/sin x (2)
=(((sin kx)^2)+1/2(cos(-2kx)-cos(2kx+2x))/sin x
=(((sin kx)^2)+1/2(cos(2kx)-cos(2kx+2x))/sin x
=((((sin kx)^2)+1/2(1-2(sin kx)^2-(1-2(sinkx+x))^2))/sin x
=((((sin kx)^2)+1/2(2(sin kx+x))^2-2(sin kx)^2))/sin x
=((sin kx)^2+(sin kx+x))^2-(sin kx)^2)/sin x
=(sin(k+1)x)/sin x
=rhs

Therefore, true for all n.

[AxS.Sorathez]

Originally Posted by Mukade

I'll use the indentity (1) (sin x)^2= (1-cos 2x)/2
because (sin nx)^2/(sin x) is easier to work with.
and also use
(2) (sin A sin B=1/2(cos(A-B)-cos(A+B))


Set n=1

lhs=sinx
rhs=(sin x)^2/sin x
=sin x
=lhs
true for n=1

Assume n=k
sin x+ sin 3x+...+sin(2k-1)x=(sin kx)^2/(sin x)

For n=k+1

lhs= ((sin kx)^2)/sin x+sin(2k+1)x
=(((sin kx)^2)+sin(2k+1)x sin x)/sin x
=(((sin kx)^2)+1/2(cos(x-2kx-x)-cos(2kx+2x))/sin x (2)
=(((sin kx)^2)+1/2(cos(-2kx)-cos(2kx+2x))/sin x
=(((sin kx)^2)+1/2(cos(2kx)-cos(2kx+2x))/sin x
=((((sin kx)^2)+1/2(1-2(sin kx)^2-(1-2(sinkx+x))^2))/sin x
=((((sin kx)^2)+1/2(2(sin kx+x))^2-2(sin kx)^2))/sin x
=((sin kx)^2+(sin kx+x))^2-(sin kx)^2)/sin x
=(sin(k+1)x)/sin x
=rhs

Therefore, true for all n.

I had quite the love/hate relationship with mathematical induction in High School...