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Find the equation of the tangent to the curve y = \sin x at x = \frac{\pi}{3}

Question:

Find the equation of the tangent to the curve {eq}y = \sin x {/eq} at {eq}x = \frac{\pi}{3} {/eq}

The Tangent Line to a Curve:

Tangent line touches a curve at a specific point. We are given a trigonometric sine function and we need to find out the tangent line at the given point.

To find out the slope, we need to compute the derivative by using the common derivative for a trigonometric function: {eq}\dfrac{\ d }{ \ dx } ( sin x ) = \cos x. {/eq}

Answer and Explanation:

We are given: {eq}y = \sin x {/eq}


First we need to find the tangent point:

Plug in {eq}x = \dfrac{\pi}{3} {/eq} into the equation {eq}f (x) = \sin \dfrac{\pi}{3} = \dfrac{\sqrt 3}{2} {/eq}.

Hence the tangent point is {eq}( \dfrac{\pi}{3} , \dfrac{\sqrt 3}{2} ) {/eq}.


Next, compute the slope of {eq}y = \sin x {/eq}:

{eq}m=\dfrac{\ dy }{\ dx}= \dfrac{\ d }{\ dx} (\sin x)= \cos x {/eq}


Plug in {eq}x=\dfrac{\pi}{3} {/eq} then {eq}m= \cos \dfrac{\pi}{3} =\dfrac{1}{2} {/eq}


Plug in {eq}m=\dfrac{1}{2} {/eq} and the point {eq}( \dfrac{\pi}{3} ,\dfrac{\sqrt 3}{2} ) {/eq} in the equation of a tangent line, we'll get:

{eq}\dfrac{\sqrt 3}{2} = \dfrac{1}{2}\cdot \dfrac{\pi}{3}+b \Rightarrow b= \dfrac{\sqrt 3}{2} -\dfrac{\pi}{6} {/eq}


Therefore an equation of the tangent line to the curve {eq}f(x) = \sin x {/eq} when {eq}x = \frac{\pi}{3} {/eq} is: {eq}{\boxed{ y =\dfrac{1}{2}x + \dfrac{\sqrt 3}{2} -\dfrac{\pi}{6}. }} {/eq}


Learn more about this topic:

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Tangent Line: Definition & Equation

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11
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