# Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 29

The equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by $$y=(1-\sqrt{3}) \theta+\sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$

#### Work Step by Step

Since $y=2(\sin \theta+\cos \theta)$, then one can find that $y'=2(\cos \theta-\sin \theta)$ and hence the slope at $\theta=\frac{\pi}{3}$ is $m=2(\cos \frac{\pi}{3}-\sin \frac{\pi}{3})=1-\sqrt{3}.$ Now, the equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by $$y=(1-\sqrt{3}) \theta+c.$$ Since the curve and the line coincide at $\theta=\frac{\pi}{3}$, then $$2(\sin \frac{\pi}{3}+\cos \frac{\pi}{3})=(1-\sqrt{3}) \frac{\pi}{3}+c \Longrightarrow c= \sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$ Hence the equation of the tangent line at $\theta=\frac{\pi}{3}$ is given by $$y=(1-\sqrt{3}) \theta+\sqrt{3}+1-\frac{\pi(1-\sqrt{3})}{3} .$$

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