Question

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

Answer 1

`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`