I kinda like arc, ^-1 kinda makes you want to make it a reciprocal not an inverse.
well reciprocal and inverse means the same thing.........
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i dont know if you figured this out yet, but thats not how inverse functions workcos(arccot(3/4) (there is a whole series of these, I think there is a trick I am missing, there is no regular radian for 3/4)
cotangent inverse gives the angle of the cotangent with the contangent's value as its domain. use that and you can just draw a triangle and get A/H.
I am still confused, not considering cosine yet, the problem is basically cot (x) = 3/4, right? Im not quite sure for exactly cotangents work, but the only solutions I know for tangent are fractions on the basic unit circle (1, (sqrt(3)/2)/(1/2), the opposite of that, and 0). Please forgive me if I completely off, I haven't taken trig anything in over 2 years which is part of my problem.
it's asking for the cosine of the angle that makes cotangent 3/4. cotangent = A/O = 3/4.
the triangle has an angle, it can be either angle other than the 90 it really doesn't matter. its opposite sid is 4 and its adjacent side is 3.
ew trigonometry bleh
Canadians use sin^-1, if that means anything
I was wondering what arcsin was until I read Rayne's post. Then I was like "Oh, sin^-1 !"
Originally made by LM:
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what is the function that you need to plug 5 into to get the original function back