Question

Find the equation of the tangent line to the curve y = 4sin x at the point (pi/6, 2) .

Answer 1

Solution

Step 1

$\begin{gathered} y=4sinx \\ \\ \frac{dy}{dx}\text{ = 4cosx} \\ \\ \frac{\pi}{6}\text{ = 30} \\ \\ m\text{ = 4cos30} \\ \\ m\text{ = 2}\sqrt{3} \end{gathered}$

Step 2

$\begin{gathered} y\text{ -2 = 2}\sqrt{3}\text{ \lparen x-}\frac{\pi}{6}) \\ \\ y\text{ - 2 = 2}\sqrt{3}x\text{ - }\frac{\pi\sqrt{3}}{3} \\ \\ y\text{ = 2}\sqrt{3}x\text{ + 2 - }\frac{\pi\sqrt{3}}{3} \\ \\ b\text{ = 2 - }\frac{\pi\sqrt{3}}{3} \end{gathered}$$\begin{gathered} m\text{ = 2}\sqrt{3} \\ \\ b\text{ = 2 - }\frac{\pi\sqrt{3}}{3} \end{gathered}$