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
Shwetank Mauria
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#

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