Write an algebraic expression for cos(sin^-1 x), cosine of inverse sine x

an algebraic expression for cos(sin^-1 x)

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[truth]

you saved us.! the easiest explanation by far!

[selwannissan]

Very clear and easy to follow, thank you!!

[mbpro]

Thanks to you, I was able to optimise a trilateration fonction. This "Y = A * cos(asin(abs(Z) / A))" turned into this "Y = sqrt(A^2 - abs(Z)^2)". I've been thinking all night about it, so I was quite happy to find this video :)

[TheSuperkiller]

Thank you very much. Our teacher teaches us almost nothing, and then gives us the most complicated homework in life...

[hailiebaker]

Awesome video. Best explanation out there! Even 5 yrs later 🙏🏻

[hatch_]

You just saved my trig grade. Thank you

[giselagarcia]

I have a test soon on this and you saved me thanks!!

[emmalarson]

This made it so much clearer! Thank you :)

[mvennela]

Thank u very much it made my solution easier.

[seppevangestel]

Thank you very much sir, you have helped me a lot for my finals

[lifewithjulia]

Thank you so much!! This helped me when my teacher couldn’t!

[mckennak]

thank you so much! I appreciate you making this video more than words can express :):):)

[lazlo]

You are the best! Can’t thank you enough!

[jasperjohnson]

You did a good job. It really helped me. Thank you

[studyzone]

Yes but you make big solution this math

[ishratjahan]

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[zza]

I found with an other way :
let g being the inverse function of f
[we know that] f (g(x)) = x
[derivate both] f ' (g(x) * g '(x) =1
[we end up with] g'(x)=1/f '(g(x))
we also know that the derivate of sin^-1(x) is equal to 1/√(1-x^2)
using the formula "g'(x)=1/f '(g(x))" (f=sin(x) and g=sin^-1(x)) we end up with 1/√(1-x^2) = 1/cos(sin^-1(x)
so we can conclude that √(1-x^2) = cos(sin^-1(x)

notice that sin(cos^-1(x) is also equal to √(1-x^2) =)

[chrissyf]

i don't usually comment on but this was brilliant and very helpful

[robydomp]

i should integrate dx
i divided in two integral : Sx^3*sqrt(4-x^2)*cos(x/2) dx + 1/2Ssqrt(4-x^2)dx
As sqrt(4-x^2) leads to pitagorean theorem,
i got Adj. sqrt(4-x^2) Opp. x and Hyp. 2 .
So i substitute in sin and cos identities as sinT =x/2, cosT =sqrt(4-x^2)/2, x=2sinT, dx=-2cosTdT
i obtain
and i can't go on
how do I evaluate that cos(sinT)? I need to do another kind of substitution?

[ashleylee]

Thank you, that was an awesome explanation.