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+--- Thread: Trig >_>. (/showthread.php?tid=17835)
Trig >_>. - Corn - 2009-10-20
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.
Trig >_>. - Russt - 2009-10-20
ClawofBeta Wrote: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.
ClawofBeta Wrote: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 CAnd 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.
Trig >_>. - Corn - 2009-10-20
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.
Trig >_>. - Russt - 2009-10-20
cause we assumed in the beginning that the measure of A was arcsin 12/13.
Trig >_>. - Corn - 2009-10-20
Oh man, I get it.
Thanks a lot. I would have never seen that
.
Trig >_>. - Russt - 2009-10-20
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.