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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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