MyThreeSons Posted January 11, 2009 Share Posted January 11, 2009 I haven't had a child take trig yet, so I'm a little rusty. I keep going in circles; I'm sure I'm making it tougher than it really is. There are two brothers hoping I've got this figured out by the time I get to church tonight. tan (2A) = 2 sin (A) find all positive angles A under 360 for which the above equation is valid. TIA; some explanation would be nice. Quote Link to comment Share on other sites More sharing options...
kiana Posted January 11, 2009 Share Posted January 11, 2009 tan (2A) = sin (2A)/cos(2A), so sin (2A)/cos(2A) = 2 sin A. Multiply both sides by cos (2A), you get sin (2A) = 2 sin A cos (2A). Using the double-angle formulas, you get 2 sin A cos A = 2 sin A (cos^2 A - sin^2 A) Subtracting 2 sin A cos (2A) from both sides (so we can factor), we get 2 sin A cos A - 2 sin A (cos^2 A - sin^2 A) = 0. Factoring, we get (2 sin A)(cos A - cos^2 A + sin^2 A)= 0. So either sin A = 0, or cos A - cos^2 A + sin^2 A = 0. sin A = 0 has the solutions of 0 and 180. (not sure if you want to count 0 or not, since you said positive it wouldn't strictly be included) To solve the second factor, recall that sin^2 A = 1 - cos^2 A. Substituting this in, we see that -2 cos^2 A + cos A + 1 = 0. This is a quadratic polynomial in cos A which may be solved by any method you choose; the roots are 1 and -1/2. cos A = 1 has a solution only at 0, and cos A = -1/2 has solutions at 120 and 240. The solutions are 120, 180, and 240 degrees. (and 0, depending on whether you include that or not. I wouldn't, since it says positive.) Quote Link to comment Share on other sites More sharing options...
MyThreeSons Posted January 11, 2009 Author Share Posted January 11, 2009 thank you!! I'm printing this out so I can work thru it. Can I go on record (again) as saying I REALLY dislike the Abeka upper math textbooks? Quote Link to comment Share on other sites More sharing options...
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