Is sin^(-1)cosx equal to 1?
HI!!
\[\sin^{-1}\cos(x)\]
Hey there!
\[\sin^{-1}(\cos(x))\] like that?
Yeah, that is it. I'm thinking it is one since cos and sin are integrals/derivatives of each other, and they both equal one in the double angle identity
no , pretty sure it is not one
i believe it is some kind of saw tooth function
mmm, okay. The part where this comes from is this other equation... \[\int\limits_{cosx}^{0}\frac{ dt }{ \sqrt{1-t^2} }\]
hard to say what it is really \[\sin^{-1}(\cos(0))=\sin^{-1}(1)=\frac{\pi}{2}\]for example
mmm, yeah, I figured. I was trying to see if maybe they would cancel each other in some way through algebra
can you post the entire quesition?
at this point you get exactly what you wrote as an answer, a function, not a number
\[\sin^{-1}(0)-\sin^{-1}(\cos(x))\]
The question is: Find dy/dx of the following equation: \[y=\int\limits_{cosx}^{0}\frac{ dt }{ \sqrt{1-t^2} }\]
aka \[-\sin^{-1}(\cos(x))\]
they dont give me much to work with
lol i knew it !!
you are doing too much work
the fundamental theorem of calculus says the derivative of the integral is the integrand
so first is \[\int_0^{\cos(x)}\frac{1}{\sqrt{1-t^2}}dt\]
then replace \*t\) by \(\cos(x)\) and by the chain rule, multiply by the derivative of cosine
oops meant first is \[-\int_0^{\cos(x)}\frac{1}{\sqrt{1-t^2}}dt\]
then the derivative is \[-\frac{1}{\sqrt{1-\cos^2(x)}}\times (-\sin(x))\]
you can clean it up a lot, like get rid of the two minus signs, and replace the denominator by \(|\sin(x)|\)
mmm, okay, thank you very much! I just didn't read the question and went straight into integrating lol. I really appreciate your help!
\[\color\magenta\heartsuit\]
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