Date: 2/1/96 at 14:27:44 From: Anonymous Subject: Trigonometric Formulas My question concerns trigonometric formulas. How do I show how the double angle formulas are derived from the compound angle formulas? I think these are the compound angle formulas: Note: / stands for theta 1. cos(/1 + /2) = cos /1 cos /2 - sin /1 sin /2 2. sin(/1 + /2) = sin /1 cos /2 + cos /1 sin/2 Here are the double angle formulas: 1. cos 2/ = cos^2 / - sin^2 / 2. sin 2/ = 2 sin / cos / I began by trying the compound formula #2 sin(/1 + /2) =( sin /1 cos /2) +( cos /1 sin/2) if you multiply the right side you get = sin 2/ = sin / cos / + 2sin / + 2 cos / +cos/ sin / here is where I'm stuck... am I on the right track? How do I derive the double angle formulas from the compound angle formulas?? Thanx
Date: 2/1/96 at 16:51:40 From: Doctor Syd Subject: Re: Trigonometric Formulas Hello! Thanks for writing! I think you'll be happy to learn that deriving the double angle formula from the addition formulas isn't as difficult as you may have thought. Now, I'm not exactly sure what you were doing when you took the compound formula #2 and "multiplied by the right side." But, maybe I can give you some hints that will lead you in the right direction. Let's work through the first addition formula (the one for cosine), and then you can try to work through the second on your own. Let t = theta. Remember that t + t = 2t. So, what is the formula for cos (2t)? Well, cos (2t) = cos (t + t), right? And, from the addition formula, plug in t for theta 1 and t for theta 2 to get: cos (t + t) = (cos t)(cos t) - (sin t)(sin t) But, (cos t)(cos t) = (cos t)^2 = cos^2 t Similarly, (sin t)(sin t) = sin^2 t So, we get that cos (2t) = cos (t + t) = cos^2 t - sin^2 t Thus, we've shown cos (2t) = cos^2 t - sin^2 t, the double angle formula. See if you can follow similar steps to prove the double angle formula for sine! Good luck, and write back if you have any more questions. -Doctor Syd, The Math Forum
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.