inverse trig.

**iiharthero** · November 3rd 2009, 08:05 PM

If siny=x and pi/2<y<pi, find dy/dx in terms of x

thank you for any help given.

**purebladeknight** · November 3rd 2009, 09:39 PM

hi im new, today was my first day being introduced into inverse trig. please tell me if i am incorrect as it would benefit both of us.

If siny=x and pi/2<y<pi, find dy/dx in terms of x

y is an angle and x is a ratio.

siny=x, pi/2 < y < pi, sin is positive in this domain.
so x'= cosy

**mr fantastic** · November 3rd 2009, 10:04 PM

Originally Posted by iiharthero

If siny=x and pi/2<y<pi, find dy/dx in terms of x

thank you for any help given.

$\frac{dx}{dy} = \cos y \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y}$ .

But $\sin y = x \Rightarrow \cos y = -\sqrt{1 - x^2}$ for $\frac{\pi}{2} < y < \pi$ . Therefore ....

**purebladeknight** · November 4th 2009, 01:43 AM

Originally Posted by mr fantastic

$\frac{dx}{dy} = \cos y \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y}$ .

But $\sin y = x \Rightarrow \cos y = -\sqrt{1 - x^2}$ for $\frac{\pi}{2} < y < \pi$ . Therefore ....

hi, please explain if it is possible how $\cos y = -\sqrt{1 - x^2}$ for $\frac{\pi}{2} < y < \pi$ .

**mr fantastic** · November 4th 2009, 02:46 AM

Originally Posted by purebladeknight

hi, please explain if it is possible how $\cos y = -\sqrt{1 - x^2}$ for $\frac{\pi}{2} < y < \pi$ .

$\sin^2 y + \cos^2 y = 1$ and cosine is negative in the second quadrant.

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