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

Jan 11, 2017

$y + \frac{\sqrt{3}}{2} x - \frac{\pi \sqrt{3}}{12} - \frac{7}{4} = 0$

#### 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 $y = f \left(x\right) = 2 - {\sin}^{2} x$ and we are seeking tangent at $x = \frac{\pi}{6}$, it is essentially seeking tangent at $\left(x , f \left(x\right)\right)$ i.e.

(pi/6,(2-sin^2(pi/6)) or (pi/6,(2-(1/2^2)) i.e. $\left(\frac{\pi}{6} , \frac{7}{4}\right)$

For slope, as $f \left(x\right) = 2 - {\sin}^{2} x$, $f ' \left(x\right) = - 2 \sin x \cos x$

and slope at $\left(\frac{\pi}{6} , \frac{7}{4}\right)$ is $f ' \left(\frac{\pi}{6}\right) = - 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = - \frac{\sqrt{3}}{2}$

Using point slope form of equation $y - {y}_{1} = m \left(x - {x}_{1}\right)$

equation of tangent is $y - \frac{7}{4} = - \frac{\sqrt{3}}{2} \left(x - \frac{\pi}{6}\right)$ or $y + \frac{\sqrt{3}}{2} x - \frac{\pi \sqrt{3}}{12} - \frac{7}{4} = 0$