graphing-functionstrigonometry
How do I sketch the graph of
$Y = \sin(3x) + \sin (x)$?.
I realize the amplitude is $1$, but am not sure. I found the period to be $2\pi$. My problem is that I don't know if this graph will be similar to the graph of $Y = \sin (x)$. I want to know how the graph will be like.
Expanding on the comment by @Andre, graphs of such nature are found (physically) in applications like Vibrations and Dynamics. Think of $\sin(x)$ as a graph exhibiting the Oscillatory motion of a particle. The usual form of such equations is $A\sin(\omega t + \alpha) \text{ where }\omega t = \theta$, A is amplitude and $\alpha$ is phase lag. We have $A\sin(\theta)$.
If A is constant, we have a magnified (or diminished) sine curve. This is for a particle with an initial perturbation but 0 friction (so there is no damping).
For an unstable particle without damping, the amplitude goes on increasing with time. So, as $t$ increases, $\omega t$ increases, $\theta$ increases and also, A increases. You have the graph for $x \sin(x)$ which looks like:
.
At any point of time, the amplitude of the sine wave is in relation to the y = x and y = -x guiding lines as you can find in the comments.
First, draw the graph for $x>0$. This will be the same graph than $\tan x$. Then, since $\tan |-x|=\tan |x|$, the graph of the function will be symmetric w.r.t the axis $x=0$. Hence, you can draw the part $x<0$ just by using the symmetry of the function.
Best Answer
A simple method to sketch periodic functions like this is:
1) find all zeroes ($x$ for which $Y=0$) in the first period
$$ \sin(3x) + \sin(x) = 0 $$ $$ x = n\pi,\, x=n\pi - \frac{\pi}{2}, \quad n\in\mathbb{Z} $$ So $Y$ crosses the x-axis at $x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ for $x \in [0,2\pi]$
Edit How to find zeros: $$ \begin{align*} & \sin(x) = -\sin(3x) \\ & x = \pi + 3x + 2\pi n_1, \quad n_1 \in \mathbb{Z} \quad \# \sin^{-1}\text{ of LHS and RHS} \\ & x = 2\pi n_2 - 3x, \quad n_2 \in \mathbb{Z} \\ & \text{and solve for } x \end{align*}$$
2) find all critical points ($x$ for which $Y'=0$ or is undefined) in the first period
$$ \frac{dY}{dx} = 3\cos(3x) + \cos(x) = 0 $$ $$ x = n\pi - \frac{\pi}{2}, 2n\pi \pm 2\tan^{-1}\left(\sqrt{5\pm 2\sqrt{6}} \right), \quad n \in \mathbb{Z}$$
Edit How to find zeros: $$ \begin{align*} &\cos(x) + 3\cos(3x) = 12\cos^3(x) - 8\cos(x) = \cos(x)(3\cos^2(x)-2)=0 \\ &\cos(x)=0 \implies x=\frac{\pi}{2}+\pi n, n\in\mathbb{Z} \text{ or} \\ &3\cos^2(x)=2 \\ &\cos(x)=\pm\sqrt{\frac{2}{3}} \\ &\text{ and take the in inverse cosine of each side} \\ &x=\pm\cos^{-1}\left(\sqrt{\frac{2}{3}}\right) + 2\pi n \text{ or}\\ &x= \pm\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)-\pi\right) + 2\pi n, n \in \mathbb{Z} \\ \end{align*} $$
Where the above inverse cosine is equivalent to the inverse tangent expression I showed in my answer.
so $Y$ has critical points at $(0.615, \,1.54),\, (\frac{\pi}{2}, \,0),\, (2.526,\,1.54),\, (\pi,\,0),\, (3.757,\,-1.54),\, (\frac{3\pi}{2},\, 0),\,\text{ and } (5.668,\, -1.54)$
3) find concavity by looking at the sign of the second derivative (the sign of $Y''$). This, however, is not really necessary for a periodic function because you can just look at the intervals between, which the first derivative changes sign.
and you should get a graph like below, which repeats every $2\pi$