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Using Trigonometric Identities


Date: 05/31/99 at 15:38:40
From: Shannon
Subject: Trigonometry

Prove:
sin(4x) = 2sin(2x) - 8sin^3(x)cosx


Date: 05/31/99 at 18:57:45
From: Doctor Sorelle
Subject: Re: Trigonometry

Dear Shannon,

Have you tried to solve this problem yet? It's always nice to see what 
things you've tried to do so that we can help you see where you're 
going wrong. And, of course, you'd never send something into us 
without trying it first, right? 

Let's see if I can help...

There are lots of different ways to approach this sort of problem. You 
could try to make the 'messy' side look 'neater' or the 'neater' side 
look 'messier' or you could try to move things from side to side and 
change things until you got one of those nice math truths like 1 = 1.

I generally go for the side that looks compact and try to change it 
until it looks like the other side. I NEVER move things from one side 
to the other; it just confuses me too much. You don't have to agree 
with that, I'm just telling you how I do it.

In this problem, I'd probably work with the left side, sin(4x), and 
try to make it look like the right side. I'd think, what can I do with 
sin(4x) to make it less compact?" Then I'd look at the other side to 
give me clues as to what sort of thing I should do.

   sin(4x) = 2sin(2x) - 8(sin(x))^3(cos(x))

What do you notice?

The first thing I'd probably notice is that on the right side (what 
I'm trying to get the left side to look like) none of the sin or cos 
functions have (4x) in them. I'd figure that I should probably try to 
break that up. I'd think of what identities I know that could help 
me.

I only know one, though you might know more.

   sin(2A) = 2sin(A)cos(A)

If you make A equal to 2x, you'll find that you can change the left 
side to 2sin(2x)cos(2x), which would give you

   2sin(2x)cos(2x) = 2sin(2x)-8(sin(x))^3(cos(x))

Now step back again. What have we accomplished? Well, we have a 
2sin(2x) on both the left side and the right side and we don't want to 
do anything else with that. That means that the only thing left to 
play with on the left side is the cos(2x).

I know 3 equations for cos(2x) (again, you might know more or less):

   cos(2A) = cos(A)^2 - sin(A)^2
   cos(2A) = 2(cos(A))^2 - 1
   cos(2A) = 1 - 2(sin(A))^2

Look at the left and the right side carefully and think about which 
one it would be best to use. If you can't figure it out, try all 
three; it won't take too long. Remember that sin(2A) = 2sin(A)cos(A), 
you might need that again.

I bet you can take it from here. Good luck!

- Doctor Sorelle, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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