# trigonometry

• Jul 31st 2009, 11:17 PM
Rose Wanjohi
trigonometry
it is given that y=sin^-1,that is, y=arc sinx. by expressing x as a function of y, determine dx/dy and hence determine dy/dx as a function of x only(Cool)
• Jul 31st 2009, 11:49 PM
Chris L T521
Quote:

Originally Posted by Rose Wanjohi
it is given that y=sin^-1,that is, y=arc sinx. by expressing x as a function of y, determine dx/dy and hence determine dy/dx as a function of x only(Cool)

Note that $\displaystyle y=\arcsin x\implies\sin y =x$

Now, differentiate w.r.t. y to get $\displaystyle \cos y=\frac{\,dx}{\,dy}$.

From here, it's evident that $\displaystyle \frac{\,dy}{\,dx}=\frac{1}{\cos y}$

I leave it for you to show that if $\displaystyle \sin y=x$, then $\displaystyle \cos y = \sqrt{1-x^2}$.

Thus, it follows that if $\displaystyle y=\arcsin x$, then $\displaystyle \frac{\,dy}{\,dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$

Does this make sense?
• Jul 31st 2009, 11:53 PM
alexmahone
Quote:

Originally Posted by Rose Wanjohi
it is given that y=sin^-1,that is, y=arc sinx. by expressing x as a function of y, determine dx/dy and hence determine dy/dx as a function of x only(Cool)

$\displaystyle y=arcsin x$

$\displaystyle x=sin y$ -------- (1)

$\displaystyle \frac{dx}{dy}=cos y$

$\displaystyle \frac{dy}{dx}=\frac{1}{cos y}$

Using (1), $\displaystyle cos^2y=1-x^2$

$\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$
• Aug 1st 2009, 01:11 AM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/t...arcsinDiff.png

This isn't quite to express x as a function of y - I'll put a balloon version of that further down the post (later)... but this here is the picture I find best of all for this problem. .

As usual, straight continuous lines differentiate downwards (integrate up) with respect to the explicit variable (x or theta) and the straight dashed line similarly but with respect to the dashed balloon expression.

For enlightenment, work from the top left corner - the given expression arcsin x. Then consider arcsin as the outer function of a chain rule differentiation...

http://www.ballooncalculus.org/asy/chain.png

If the inner function is sin, then the derivative with respect to sin will be multiplied by the derivative of sin...

http://www.ballooncalculus.org/asy/t...rcsinDiff1.png

But then arcsin(sin(theta)) is of course theta, and the derivative of that is 1, so the whole of the bottom of the chain-rule shape wants to come to 1, so we know it must be...

http://www.ballooncalculus.org/asy/t...rcsinDiff2.png

Which we just re-write using Pythag and then - swapping sin theta back for x - we finish at the bottom left corner...

_______________________________

And if you do need to express x as a function of y...

http://www.ballooncalculus.org/asy/dXdY/arcsinDiff.png

... and proceed as per the other 2 posts.

Don't integrate - balloontegrate!

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