Find the equation of the tangent to the curve at the point where .

MATHEMATICS · Single correct · CBSE · Mathematics · Limits and Derivatives · Limits and Derivatives

Problem

Find the equation of the tangent to the curve

y = \sin x

at the point where

x = \dfrac{\pi}{4}

. (a)

y - \dfrac{\sqrt{2}}{2} = \sqrt{2}\left(x - \dfrac{\pi}{4}\right)

(b)

y - \dfrac{\sqrt{2}}{2} = \left(x - \dfrac{\pi}{4}\right)

(c)

y - \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{2}\left(x - \dfrac{\pi}{4}\right)

(d)

y - \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\left(x - \dfrac{\sqrt{2}}{2}\right)