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Help w/ Trig!!!

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Alright, I''m reviewing some trig and I''m running into some problems. First of all, I realize that that sin((7/6)pi) in radians and sin((11/6)pi) radians equals -.5 . When I do arcsin of -.5 I get a long rational number: -.5235987756. This number however does not equal (7/6)pi or (11/6)pi so can someone please tell me what this irrational number represents? This has been really frustrating me and I''d greatly appreciate some help. I''m sure it''s just some basic idea that I''m forgetting. Thx, --BioX

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arcsin''s domain is [-Pi/2, Pi/2]. Both 11*Pi/6 and 7*Pi/6 will have the same sin as -Pi/6, which is within arcsin''s domain. Your number is -Pi/6.


Thanks Salsa!Colin Jeanne | Invader''s Realm
"I forgot I had the Scroll Lock key until a few weeks ago when some asshole program used it. It even used it right" - Conner McCloud

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x | sin x = y | arcsin y
-------------------------------------
(-11/6)PI | .5 | .523... = (1/6)PI
(-10/6)PI | .8660... | 1.047... = (2/6)PI
(-9/6)PI | 1 | 1.570... = (3/6)PI
(-8/6)PI | .8660... | 1.047... = (2/6)PI
(-7/6)PI | .5 | .523... = (1/6)PI
(-6/6)PI | 0 | 0 = 0
(-5/6)PI |-.5 |-.523... = (-1/6)PI
(-4/6)PI |-.8660... |-1.047... = (-2/6)PI
(-3/6)PI |-1 |-1.570... = (-3/6)PI
(-2/6)PI |-.8660... |-1.047... = (-2/6)PI
(-1/6)PI |-.5 |-.523... = (-1/6)PI
0 | 0 | 0 = 0
(1/6)PI | .5 | .523... = (1/6)PI
(2/6)PI | .8660... | 1.047... = (2/6)PI
(3/6)PI | 1 | 1.570... = (3/6)PI
(4/6)PI | .8660... | 1.047... = (2/6)PI
(5/6)PI | .5 | .523... = (1/6)PI
(6/6)PI | 0 | 0 = 0
(7/6)PI |-.5 |-.523... = (-1/6)PI
(8/6)PI |-.8660... |-1.047... = (-2/6)PI
(9/6)PI |-1 |-1.570... = (-3/6)PI
(10/6)PI |-.8660... |-1.047... = (-2/6)PI
(11/6)PI |-.5 |-.523... = (-1/6)PI

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HAHAH BioAgent = Nub;
This is Chad btw from the lame GameProgramming course we had a couple years ago.

Nice to see you . You gotta tell me where you hang on IRC now.

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Oh ok, that helps, but I still have a slight problem. You see I have this equation -2sin(x) - sin(x)^2 - cos(x)^2 = 0 and I''m trying to solve for x. I''ve solved it to the point that -2sin(x) = 1(which is cos(x)^2 + sin(x)^2) and then I found that x is the arcsin of -.5, which as you said is -pi/6. But why does only -pi/6 work in the equation for zero, if 7pi/6 and 11pi/6 have the same sine shouldn''t they also be roots? They don''t seem to be when I plug them in, and I don''t know why. Thanks for the help so far.

BTW: Yea chad, it''s been a long time. To be honest I don''t go on IRC that much anymore. Hell, I don''t even program that much anymore because I lack the time. I''m trying to do well in school so I can go to college, and THEN focus on programming. Although I do it whenever I can. But when I do go on IRC I probably go to the gamedev.net room. I''ll drop you a line the next time I go on trillian.

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quote:
Original post by bioagentX
Oh ok, that helps, but I still have a slight problem. You see I have this equation -2sin(x) - sin(x)^2 - cos(x)^2 = 0 and I''m trying to solve for x. I''ve solved it to the point that -2sin(x) = 1(which is cos(x)^2 + sin(x)^2) and then I found that x is the arcsin of -.5, which as you said is -pi/6. But why does only -pi/6 work in the equation for zero, if 7pi/6 and 11pi/6 have the same sine shouldn''t they also be roots? They don''t seem to be when I plug them in, and I don''t know why. Thanks for the help so far.


If someone were to ask you, "For what x is sin x = (√2)/2?", there would be an infinite number of correct answers you could give. You could say π/4, or 3π/4, or -31π/4, or any other angle whose sine was (√2)/2.

You could give multiple responses when given a single number to work with. If you were a function, though, you''d be fired for that, because functions can only have one output associated with each input.

The same problem arises when defining arcsin and arccos. This is why their ranges are constrained to [-π/2, π/2] and [0, π], respectively.

You might encounter problems asking you for only one solution, or for all solutions, or for all solutions in a certain range. Be careful to answer the question asked in the problem.

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