How do I prove this? Trig/induction/dunno question? (1 Viewer)

[SureBluff]

I'm not sure if this is an induction question or just a normal trig question. I have no clue how to start this off. (I doubt its induction)

Use the result sinAsinB = 1/2(cos(A-B)-cos(A+B)) to prove that:

sinx + sin3x + sin5x + ...... + sin(2n-1)x = ((sinnx)^2)/sinx

so totally confused.

 

[Yip]

Its induction,

When n=1,
LHS=sinx
RHS=[sinx]^2/sinx=sinx=LHS
True for n=1

Assume true for n=k
ie sinx+sin3x+.........+sin(2k-1)x=[[sinkx]^2]/sinx
Prove true for n=k+1
Aim to Prove: sinx+........+sin(2k-1)x+sin(2k+1)x=[[sin[k+1]x]^2]/sinx

By assumption,
LHS=[[sinkx]^2]/sinx+sin(2k+1)x
=[[(sinkx)^2]+sin(2k+1)xsinx]/sinx
=[[(sinkx)^2]+1/2[cos(x-2kx-x)-cos(2kx+2x)]/sinx (as sinAsinB=1/2(cos(A-B)-cos(A+B))
=[[(sinkx)^2]+1/2[cos(-2kx)-cos(2kx+2x)]/sinx
=[[(sinkx)^2]+1/2[cos(2kx)-cos(2kx+2x)]/sinx
=[[[(sinkx)^2]+1/2[1-2(sinkx)^2-(1-2[sinkx+x)]^2)]/sinx
=[[[(sinkx)^2]+1/2[2[sinkx+x)]^2-2(sinkx)^2]]/sinx
=[(sinkx)^2+[sinkx+x)]^2-(sinkx)^2]/sinx
=(sin[k+1]x)/sinx=RHS

Induction statement
True for all natural n by Mathematical Induction

 

[Riviet]

That is a fairly hard induction question, I've done it myself and found it quite challenging.

 

[gman03]

sinx (sinx + sin3x + sin5x + ...... + sin(2n-1)x)

= sinx sinx + sinx sin3x + sinx sin5x + ...... + sinx sin(2n-1)x)

= 1/2 { ( cos 0 - cos 2x) + ( cos 2x - cos 4x) + ( cos 4x - cos 6x) + ... ( cos 2[n-1]x - cos 2nx) }

= 1/2 (cos 0 - cos 2nx)

= 1/2 (cos [nx - nx] - cos [nx + nx])

= sinnx * sinnx

= sin^2 nx

So for nonzero sinx ,

(sinx + sin3x + sin5x + ...... + sin(2n-1)x) = sin^2 nx / sinx

I don't think induction is necessary unless explicitly asked otherwise.

 

[SureBluff]

Hey Tim Yip, guess who this is.

Anyways, I found another way of doing it, its easier this way:


Use the result sinAsinB = 1/2(cos(A-B)-cos(A+B)) to prove that:

sinx + sin3x + sin5x + ...... + sin(2n-1)x = ((sinnx)^2)/sinx

Rearranging we get:

sinxsinx + sinxsin3x + sinxsin5x + .... + sinxsin(2n-1)x = (sinnx)^2

using product to sum we get:

LHS: 1/2(cos0-cos2x) + 1/2(cos2x-cos4x) + 1/2(cos4x-cos6x) + ... + 1/2(cos2nx-2x-cos2nx)

notice that the terms cancel out, therefore we're left with the first and last term:

= 1/2(cos0-cos2nx)
= 1/2(cos0 - (1- 2(sinnx)^2))
=1/2 - 1/2 + (sinnx)^2
= (sinnx)^2 = RHS

 

[SureBluff]

oops

didnt see gman's post.

ohh well

same idea.

 

[Yip]

Yeah, that method is way faster and more elegant, but imo i think in an exam it would be put as an induction q unless its in a 4u paper.
btw surebluff, are u lawrence wang? or another member of the canteen poker players?

 

[SureBluff]

ha

why did you say Lawrence first???

anyways, yes yes I am

holding tournaments during the holidays.

 

[Riviet]

Only prove something by induction if the question specifically asks you to do so.

 

[Sparcod]

ha

why did you say Lawrence first???

anyways, yes yes I am

holding tournaments during the holidays.

Hey Lawrence
World Poker Series.

 

[PhillipsNormo]

I just looking at these comments whilst doing the exact question in 2022 now (1/01/2022), when you guys did the HSC before I was even born.