Question

find the equation of the tangent line to the curve ( y = 4 sin x ) at the point ( left( \frac { pi } { 6 }, 2
ight) ).
the equation of this tangent line can be written in the form ( y = m x + b ) where
( m = )
and ( b = )
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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:

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