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/
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