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