Prove that: cos(sin^-1(x)) = sqrt (1-x^2)
Let \(a=\cos(\sin^{-1} x)\) Then \(a^2+\sin^2(\sin^{-1}x)=1\) But remember \(\sin(\sin^{-1}(x))=x\) Hence \(a^2+x^2=1\) \(a^2=1-x^2\) \(a=\sqrt{1-x^2}\) Therefore \(\cos(\sin^{-1}(x)=\sqrt{1-a^2}\)
I mean \(\cos(\sin^{-1}(x))=\sqrt{1-x^2}\)
Do you mind if I ask you another question
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hello wathmath!
lots of posts tonight
wow you already halfway amistre64 medals :D
I don't know I am not as excited as before in answering questions :D
well i have to agree. getting old, but my latex is improving by leaps and bounds
and no amistre is way ahead. i will never catch up
use \ ( \ ) instead of \ [ \ ] to get inline latex, that will make your post more concise :)
yo satellite can you go back to this question that you were helping me with? If sin(x) = 1/3 and sec(y) = 5/4 , where x and y lie between 0 and pie/2, evaluate the expression cos(x+y).
I asked you about how the pathagoras part you did i was a little confused..can you show that work?
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