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

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.

• You may find it useful to note that $\sin(A)+\sin(B)=2\sin((A+B)/2)\cos((A-B)/2)$. (For instance, to calculate roots) – preferred_anon Mar 11 '17 at 15:31
• One way to do it is to sketch both $y=\sin(3x)$ and $y=\sin(x)$ on the graph, and then sketch the values of the sum of the two functions. – projectilemotion Mar 11 '17 at 15:35
• Also find some interesting special points. Like all of the zeros, max, and mins that you can. Plot those individual points first and then smoothly connect them. – Brick Mar 11 '17 at 15:38
• "I realize the amplitude is $1$": no, think of the value of the function at $x=\pi/6$. – Yves Daoust Mar 11 '17 at 15:40
• You could expand $sin(3x)$ in terms of $sin(x)$ as $sin(2x+x)$, and use addition formulae to expand this. This will give a cubic equations which you can graph – Sumant Mar 11 '17 at 15:40

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 $2\pi$

• What's the idea behind getting the zeroes of X? How did X=nπ? – Ashalley Samuel Mar 12 '17 at 13:02
• Ohk I get it now. 2x=2nπ which implies X=nπ – Ashalley Samuel Mar 12 '17 at 13:14
• @AshalleySamuel $\sin x$ is periodic and crosses the x-axis infinitely many times, so it has infinite zeroes. In the case of your function, it has zeroes at integer multiples of $\pi$ $(\pi, 2\pi, 3\pi, ...)$ and also $\frac{\pi}{2}$ less than integer multiples of $\pi$ $(\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...)$ – Dando18 Mar 12 '17 at 13:15
• I see. This is how I solved the X. Sin(3x)=Sin(-x)> 3x=(-x+2nπ) and 3x={(π+x)+2nπ} so eventually you will have X=nπ/2 and X=(π/2+nπ). How different is this from yours? – Ashalley Samuel Mar 12 '17 at 13:31
• @AshalleySamuel see my edit – Dando18 Mar 12 '17 at 13:51

In this particular case, you can bypass a complete study using a trick: $\sin3x$ is a "compressed" version of $\sin x$, as if the $x$ axis had been shrunk by a factor $3$.

You can plot both functions on the same graph and perform a graphical addition. Then, you should understand the hint below.

Hint: MW.

One way to do this is like this:

Sketch both $y=\sin(3x)$ and $y=\sin(x)$. Note that the period of $y=\sin(x)$ is $2\pi$, therefore the period of $y=\sin(3x)$ is $\frac{2\pi}{3}$. Therefore, we can sketch both graphs. Here, the blue curve is $y=\color{blue}{\sin(x)}$ and the red curve is $y=\color{red}{\sin(3x)}$: The black points are a result of graphical addition (Summing up both functions). Since you are adding both functions, you are essentially plotting: $$y=\color{red}{\sin(3x)}+\color{blue}{\sin(x)}$$

I've done the first few for you. Continue this process, and look for a pattern. Connect the black points with a smooth curve.

• Pls with this approach, how will I know when the black dotted lines should cross the X axis and when it will not? – Ashalley Samuel Mar 12 '17 at 14:53
• When the sum of the two functions is equal to $0$, it will cross the $x$-axis. For example, at $x=\frac{\pi}{2}$, the $y$-coordinates of both functions are $1$ and $-1$, hence the sum is equal to $0$. Thus, $y=\sin(3x)+\sin(x)$ will intersect with the $x$-axis at $\left(\frac{\pi}{2},0\right)$. – projectilemotion Mar 12 '17 at 16:50