How do i find the equation of the tangent line? Find the function of the tangent line to $$f(\theta)=5\left(\cos(\pi/3)\cdot\sin(\theta)\right)$$ at $\theta=2\pi/3$
Please show steps and explain. I need to understand this for an upcoming exam.
 A: $$f(\theta)=5[\cos(\pi/3)*\sin(\theta)]$$ 
We want to find $y = mx + b$
$$m = f'(\theta)=\frac{5}{2} Cos(\theta) = -\frac{5}{4} ~\text{when}~ \theta = (2\pi)/3$$
$$b = f(2\pi/3) = \frac{5 \sqrt{3}}{4}$$
So, we get 
$$y = -\frac{5}{4} x + \frac{5 \sqrt{3}}{4}$$
A: Compute the derivative of $f(\theta) = 5[\cos(\pi/3)\sin \theta]$, using, say, the product rule.
Then evaluate $f'$ at $\theta = 2 \pi/3$. Then $f'(2\pi/3) = m$ where $\;m = $ the slope of the line tangent to $f(\theta)$.
Use the fact point-slope form of an equation to determine the precise equation of the line:
$$y - y_0 = m(x - x_0)$$
where $(x_0, y_0)$ is a point known to be on the line. 
The point we know is on the line is $\left(2\pi/3, f(2\pi/3)\right)$, so $x_0 = 2\pi/3,\;\;y_0 = f(2\pi/3)$
So the equation of your line will look like this: $$y - f(2\pi/3) = m[x - f(2\pi/3)] = y - f(2\pi/3) = f'(2\pi/3)(x - f(2\pi/3))$$
where $f(2\pi/3)$ is the value of your original equation $f(\theta)$ evaluated at $\theta = 2\pi/3$,
and $f'(2 \pi/3)$ is the value of the derivative of $f(\theta)$ evaluated at $\theta = 2\pi/3$.
