You are right, but because y = arccos(x), we know that y lies in [0,pi]. For that range, sin(y) is never negative, so we only use the + answer. It would have been nice if Khan pointed this out.
@@kredziigaming981 Michele's comment is about the step at 1:37, where Khan rewrites sin y as sqrt(1-(cos y)^2). With sqrt I mean "square root", with ^2 I mean "to the power of 2". Khan does this because (sin y)^2 + (cos y)^2 = 1, perhaps you remember this from trigonometry. From this it follows that (sin y)^2 = 1 - (cos y)^2. Then either sin y = sqrt(1-(cos y)^2) or sin y = -sqrt(1-(cos y)^2)... Khan only uses the first, positive, solution and not the second, negative, solution. That's what Michele's question was about. My point was that the negative solution is not relevant because y is given as the inverse cosine of x (Khan writes inverse cosine as cos^-1, but it's also written as arccos, which is how I wrote it). The inverse cosine only gives values between 0 and pi, and sine for these angles is positive.
As always, Khan Academy has the go-to video for understanding.
Bless your soul. I’ve been trying to find this prove
So d/dx of cos^-1(x)+sin^-1(x) is equal to zero!
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very helpful
Would you start writing big words I love your works but I have problem with my eyes am finding difficulties in following.i will appreciate
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isn't sin(y) also equal to -sqrt(1-cos^2y)? Why did we pick the principal square root, being the negative a totally legit option?
You are right, but because y = arccos(x), we know that y lies in [0,pi]. For that range, sin(y) is never negative, so we only use the + answer. It would have been nice if Khan pointed this out.
@@kredziigaming981 Michele's comment is about the step at 1:37, where Khan rewrites sin y as sqrt(1-(cos y)^2). With sqrt I mean "square root", with ^2 I mean "to the power of 2". Khan does this because (sin y)^2 + (cos y)^2 = 1, perhaps you remember this from trigonometry.
From this it follows that (sin y)^2 = 1 - (cos y)^2. Then either sin y = sqrt(1-(cos y)^2) or sin y = -sqrt(1-(cos y)^2)... Khan only uses the first, positive, solution and not the second, negative, solution. That's what Michele's question was about.
My point was that the negative solution is not relevant because y is given as the inverse cosine of x (Khan writes inverse cosine as cos^-1, but it's also written as arccos, which is how I wrote it). The inverse cosine only gives values between 0 and pi, and sine for these angles is positive.
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So that means arcsin(x) really is -arccos(x) or is -arccos(x) + some constant?
why can't you just use chain rule? I don't understand
very very helpful