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Math Help - inverse trig.

  1. #1
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    inverse trig.

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

    thank you for any help given.
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  2. #2
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    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
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  3. #3
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    Quote Originally Posted by iiharthero View Post
    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 ....
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    Quote Originally Posted by mr fantastic View Post
    \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.
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  5. #5
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    Quote Originally Posted by purebladeknight View Post
    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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