anyone want to help with a trig problem?

[MyThreeSons]

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.

[kiana]

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.)

[MyThreeSons]

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?