1. inverse trig.

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

thank you for any help given.

2. 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

3. 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 ....

4. 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 $\frac{\pi}{2} < y < \pi$.

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.