# [Math] Sketching the graph of $Y=\sin(3x)+\sin(x)$

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.

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 $2\pi$ 