How do you find the equation of a tangent line to an equation: $\sin$ in it? This is the question I need to answer, but I don't know how to.
Find an equation of the tangent line to $y=10\sin(x)$ at $x=\pi$.
 A: Hint :
The slope of the tangent is equal to the derivative of $$f(x)=10\sin(x)$$ at $x=\pi$.
A: HINTS:
What is the slope? What is the x-displacement? What is the equation of a straight line? and where is your work?
A: If $y=10\sin(x)$ then the derivative of $y$ with respect to $x$ is given by: $$\cfrac{\mathrm{d}y}{\mathrm{d}x}=10\cos x$$
This is the slope or gradient of the line we seek. 
At $x=\pi$ 
$$\cfrac{\mathrm{d}y}{\mathrm{d}x}=10\cos \pi=-10$$
the General equation of a straight line is given by
$$y-y_1=\cfrac{\mathrm{d}y}{\mathrm{d}x}(x-x_1)\tag{1}$$ where ($x_1, y_1$) are known points that lie on the line:
So when $x=\pi$, $y=10\sin(\pi)=0$
Substituting $(x_1,y_1)=(\pi,0)$ into $(1)$
$$\implies y-0=-10(x-\pi)$$
$$\implies \color{red}{y=-10(x-\pi)}$$
A: Notice, we have $$y=10\sin x$$$$\implies \frac{dy}{dx}=10\cos x$$ hence the slope ($m$) of the tangent at $x=\pi$ $$m=\left[\frac{dy}{dx}\right]_{x=\pi}=10\cos \pi=-10$$ 
Now, the corresponding y-coordinate for $x=\pi$ is $$y=10\sin \pi=0$$ 
 Hence, the equation of the tangent line passing through the point $(\pi, 0)$ & having slope $m=-10$ is given by point-slope form $$y-0=-10(x-\pi)$$ $$\color{blue}{y=-10x+10\pi}$$ 
