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    4. Find the equation to the tangent to the curve  cos(2y+\pi) at (0,\frac{\pi}{4})

    Give your answer in the form  y = ax + b , a and b are constants to be found.

    My working:

     \frac{dx}{dy} = -2sin(2y+\pi)

    At  (0,\frac{\pi}{4}) y = \frac{\pi}{4}

     \frac{dx}{dy} = -2sin(\frac{2\pi}{4} + \pi)

    Tangent =  x = -2sin(\frac{3\pi}{2})y + c

    At x = 0, y =  \frac{\pi}{4}

     0 = -2sin(\frac{3\pi}{2}) \times \frac{\pi}{4} + c

     0 = -0.19905 + c

     c = 0.12905

     x = -2sin(\frac{3\pi}{2})y + 0.12905

     y = \frac{x-0.12905}{-2sin(\frac{3\pi}{2})}

     y = \frac{1}{-2sin(\frac{3\pi}{2})}x - \frac{0.12905}{-2sin(\frac{3\pi}{2})}

     y = -6.086x + 0.785

    This just didnt seem right...

    Please help!


    Once you work out dx/dy, get the reciprocal so you have dy/dx. Then subsitute y = pi/4 for the gradient of the tangent, then use the y-y1 = m(x-x1) formula using the coordinates and gradient to get the equation of the tangent.
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Updated: December 20, 2010
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