Calculas

find the tangent line of the following function Y=2sinx + sinx^2 at the points Pi/6,5/4

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asked by Bashar
  1. I think you have a typo. I'm sure you meant

    y = 2sinx + (sinx)^2
    since
    y(π/6) = 2(1/2) + 1/4 = 5/4

    So,
    y' = 2cosx + 2 sinx cosx
    at x=π/6,
    y' = 2(√3/2) + 2(1/2)(√3/2)
    = √3 + √3/2
    = 3√3/2

    Now you have a point and a slope. Dig back into your algebra I and come up with the line:

    y - 5/4 = 3√3/2(x-π/6)

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    posted by Steve

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