Trig >_>.

I can't post in the Math Help thread, or it'll be considered necroing. Darn.

Anyways, I can't figure out this trig problem.

Evaluate the following in radians. No calculators may be used.

sin (2Arcsin 12/13)

Notice the capital A in Arcsin. A lowercased arcsin is the name as Sin^-1.

The answer is either 120/109 or 120/104 (I can't read my own handwriting, goddangit). However, I would like to know how you would do this.

Thanks.

I can't post in the Math Help thread, or it'll be considered necroing. Darn.

Technically, though, you're contributing to the thread by asking a math-related question. At least if I was a mod, I wouldn't call you out on it.

Notice the capital A in Arcsin. A lowercased arcsin is the name as Sin^-1.

What? Arcsin is inverse sine no matter what case it's in o_o

Anyway, say you have a right triangle:

Code:

A
|\
| \
|__\
B   C

And say that the measure of angle A is arcsin(12/13). This means sin A = 12/13, and we know that sin is opposite/hypotenuse, so we can assume that the opposite (length BC) is 12 units and the hypotenuse (length AC) is 13 units. Then we can invoke Pythagorean Theorem, and find that side AB is 5 units.

Now use the relation sin 2x = 2 sin x cos x. We know that sin A is 12/13. Since cos is adjacent/hypotenuse, cos A is AB/BC in the above triangle, or 5/13. So:
sin 2arcsin 12/13 = sin 2A
= 2 sin A cos A
= 2 (12/13) (5/13)
= 120/169.

Presumably that's what your messy handwriting meant.

Wait. Can you explain why sin 2arcsin 12/13 = sin 2A? I got everything else, except that.

I included the sin^- 1 part because some schools don't teach arcsin, instead using sin^-1. For some reason.

cause we assumed in the beginning that the measure of A was arcsin 12/13.

Oh man, I get it.

Thanks a lot. I would have never seen that .

Well technically you don't need the triangle, you just need sin^2 + cos^2 = 1. But I think the triangle is nicer, and that's how I'd picture it when I work with inverse trig crap.