find the equation of the tangent line to the curve ( y = 4 sin x ) at the point ( left( \frac { pi } { 6 }…

find the equation of the tangent line to the curve ( y = 4 sin x ) at the point ( left( \frac { pi } { 6 }, 2 \right) ).\nthe equation of this tangent line can be written in the form ( y = m x + b ) where\n( m = )\nand ( b = )\nquestion help: video message instructor\nsubmit question jump to answer

find the equation of the tangent line to the curve ( y = 4 sin x ) at the point ( left( \frac { pi } { 6 }, 2 \right) ).\nthe equation of this tangent line can be written in the form ( y = m x + b ) where\n( m = )\nand ( b = )\nquestion help: video message instructor\nsubmit question jump to answer

Answer

Explanation:

Step1: Find the derivative of the function

The derivative of (y = 4\sin x) is (y'=4\cos x).

Step2: Calculate the slope (m)

Substitute (x = \frac{\pi}{6}) into (y'). (m = 4\cos\frac{\pi}{6}=4\times\frac{\sqrt{3}}{2}=2\sqrt{3})

Step3: Use the point - slope form to find (b)

The point - slope form is (y - y_1=m(x - x_1)), where ((x_1,y_1)=(\frac{\pi}{6},2)) and (m = 2\sqrt{3}). (y-2=2\sqrt{3}(x-\frac{\pi}{6})) (y=2\sqrt{3}x-\frac{\pi\sqrt{3}}{3}+2) So (b = 2-\frac{\pi\sqrt{3}}{3})

Answer:

(m = 2\sqrt{3}), (b=2-\frac{\pi\sqrt{3}}{3})