How do you find the equation of the tangent to the curve y = 2 - sin^2 x at the point where x = pi/6?

1 Answer
Jan 11, 2017

y+sqrt3/2x-(pisqrt3)/12-7/4=0

Explanation:

One important thing to note when finding the equation of a tangent to a curve at a given point is that the slope of the tangent is equal to the value of first derivative at that point.

As the function is y=f(x)=2-sin^2x and we are seeking tangent at x=pi/6, it is essentially seeking tangent at (x,f(x)) i.e.

(pi/6,(2-sin^2(pi/6)) or (pi/6,(2-(1/2^2)) i.e. (pi/6,7/4)

For slope, as f(x)=2-sin^2x, f'(x)=-2sinxcosx

and slope at (pi/6,7/4) is f'(pi/6)=-2xx1/2xxsqrt3/2=-sqrt3/2

Using point slope form of equation y-y_1=m(x-x_1)

equation of tangent is y-7/4=-sqrt3/2(x-pi/6) or y+sqrt3/2x-(pisqrt3)/12-7/4=0