Cal Question Show that cos(arctan(sin(arccot(x)))=sqrt{frac{x^2+1}{x^2+2}}
\[\cos(\arctan(\sin(arccot(x)))=\sqrt{\frac{x^2+1}{x^2+2}} \]
what are you kidding?
myininaya, if I had to put money on it, I'd bet you already have this solved
lol this is a question in the book i just thought it would fun
no i was just looking and seen it and thought i would give it a try
actually i bet i can do it without any calculus at all yes? you have an odd idea of fun
i was thinking that satellite
this question looks ridiculous!
i know
lol
thats why i chose it
oh i lied
i thought i could just work from the inside out but i cannot. so i guess you do ... would you like my guess?
myininaya, just post the solution already.
i just started working on it
actually pretty easy
my guess is take the derivative of both sides. see that you get the same thing. then evaluate somewhere, see that you get the same thing. that is my guess
i just got it.
just make 2 right triangles
zarkon's right, its not that bad.
i have them on my paper. but what do you do with.. oh never mind. i am a moron
last step is \[\cos(\arctan(\frac{1}{\sqrt{x^2+1}})\] if i am not mistaken
right.
Can somebody post it on a pdf?
i can write it out maybe, it is not bad
easy
start with \[\sin(\tan^{-1}(x))=\frac{1}{\sqrt{x^2+1}}\]
i gotta learn to do that. do you scan these?
i'm a scanner
with a scanner
lol
lol
got it. oh look some one posted a nice integral for you to do!
i scan the hell out of things
zarkon remember it is my favorite thing to do to make right traingles lol
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e4894120b8b8d00ebe06793
yep...fun for all ages :)
go get 'em. i'm off.
i could give this to my trig students why would you want to show this using cal?
i don't know if that is possible to use cal for this problem
doing it once is good for trig sub...but having to do it twice..i don't see a point
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