Date: 04/21/97 at 20:45:03 From: Ailsa Subject: Trigonometry Prove that (sin 50 x cos 50)/tan 50 = sin^2 40.
Date: 06/25/97 at 17:57:17 From: Doctor Sydney Subject: Re: Trigonometry Hi Ailsa, These problems can be tricky if you don't see where to start, but usually once you start, they aren't so bad. So, we are going to get you started working on this problem, and then maybe you can finish it up. Whenever I am supposed to simplify an expression with sin's, cos's, and tan's, I convert the tan's to sin/cos's, so that everything I am dealing with is in terms of sin and cos. Usually I find then that I can cancel some of the cos's and sin's or reduce in some other way. So, let's try that with this problem. 1. First break up tan into sin and cos: (sin 50 x cos 50)/tan 50 = (sin 50 x cos 50)/(sin 50/cos 50) 2. Then simplify fractions: sin 50 x cos 50/(sin 50/cos 50) = sin 50 x (cos 50/1) x (cos 50/sin50) = sin 50 x (cos^2 50/sin 50) 3. Make cancellations: (sin 50/1) x (cos^2 50/sin 50) = cos^2 50 So, now you have that (sin 50 x cos 50)/tan 50 = cos^2 50. A big part of the problem is done. Now, all you have to do is prove that cos^2 50 = sin^2 40. Can you figure out how to do this on your own? What do you know about cos^2x and sin^2x? Notice that 50 + 40 = 90. If you need more help, please write back. Good luck! -Doctors Kathryn and Sydney, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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