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can someone please help?

This problem asks to find the equation of the tangent line to $y= 3\sin(x)$ at $x= \frac\pi 3$.

I tried finding two points of the graph and finding the slope of the line by taking the derivative of $3\sin(x)$, am I in the right direction? I think I'm doing something wrong because I can't find the correct answer.

Thank you!

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1. Find the slope of the line. The slope of the line tangent to $y = 3 \sin(x)$ at $x = \pi/3$ is: $$ \left. \frac{d}{dx}\left[3\sin(x) \right] \right|_{\pi/3} = 3\cos\left(\frac{\pi}{3}\right) = \boxed{\frac{3}{2}}$$ 2. Find a point that the line goes through; since you know that this line is tangent, then there's an easy choice -- $\left(\frac{\pi}{3}, \frac{3\sqrt{3}}{2}\right)$.

3. Use the point-slope formula for a line, namely: $y - y_0 = m(x - x_0)$. Thus, putting it together, you get: $$\boxed{y = \frac{3}{2}\left(x - \frac{\pi}{3}\right) + \frac{3\sqrt{3}}{2}}$$

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hint: $m = f'(\pi/3) = 3\cos(\pi/3) = 3/2$,and use the point slope equation: $y - 3\sqrt{3}/2 = 3/2\cdot \left(x - \pi/3\right)$

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    $\begingroup$ How is this a hint? $\endgroup$ – J.Gudal Dec 8 '15 at 20:35
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Since the equation of a line can be expressed as:

$y-y_1=m(x-x_1)$, you simply need to find the gradient of the line,$m$, and the coordinates of a point on the line, $(x_1,y_1)$.

Find the value of the derivative of $y$ at $x=\pi /3$ and find the value of $y$ at this point and you shall have all that you need.

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