Rewrite $f(x) = 3 \sin (\pi x) + 3\sqrt{3} \cos (\pi x)$ in the form $f(x) = A \sin (Kx+D)$

Question

I got a question like that said "Rewrite $f(x) = 3 \sin (\pi x) + 3\sqrt{3} \cos (\pi x)$ in the form $f(x) = A \sin (Kx+D)$".

I'm inclined to think that since the periods are the same ($2$), that the amplitudes will just add up. But, I'm not sure.

I also need to know the rules for combining sinusoids with different periods. I know that when you're multiplying them, the one with the longer period acts as a sort of envelope for the one with the smaller period. But what do you do with different periods when you add them?

Thanks!

evamvid

Answer 1

This is a standard thing: $$ \begin{align} & \phantom{{}={}}3\sin(\pi x) + 2\sqrt{3}\cos(\pi x) = \sqrt{\Big(3\Big)^2+\Big(3\sqrt{3}\Big)^2}\Big(\frac{3}{\bullet}\sin(\pi x)+\frac{3\sqrt{3}}{\bullet}\cos(\pi x)\Big) \\[10pt] & = 6\Big(\frac{3}{6}\sin(\pi x)+\frac{3\sqrt{3}}{6}\cos(\pi x)\Big) = 6\Big( \frac 1 2 \sin(\pi x)+\frac{\sqrt{3}}{2}\cos(\pi x) \Big) \\[10pt] & = 6\Big(\cos\alpha\sin(\pi x) + \sin\alpha\cos(\pi x)\Big) \\[10pt] & = 6 \sin(\pi x + \alpha). \end{align} $$

I leave it to you to figure out what number $\alpha$ is.

Answer 2

Hint: $\quad\sin(a+b)=\sin a\cos b+\sin b\cos a,\qquad\dfrac12=\sin t,\qquad\dfrac{\sqrt3}2=\cos t,\qquad t=?\quad$ :-)

Answer 3

Hint: $f(x)=3\sin(πx)+3\sqrt{3}\cos(πx)$. What is $A\sin(Kx+D)$? It is $A\sin(Kx+D)=A(\sin (Kx) \cos (D) + \cos (Kx) \sin (D))$. Let $A=1$. Then, let $K=\pi$, then $\sin (D)=3\sqrt{3}$, $\cos (D)=3$, i.e., $\tan(D)=\frac{1}{\sqrt{3}}$. Can you find $D$ now?

Answer 4

The standard way to combine such functions is to use the R Formula ( http://www.oocities.org/maths9233/Trigonometry/RFormula.html) :

Specifically, if we wanna combine your function into :

$$a\sin{\theta} + b\cos{\theta} = R\sin{(\theta + \alpha)}$$

Then it is possible to compute $R,\alpha$ :

$$R = \sqrt{a^2 + b^2}, \alpha=\tan^{-1}{\frac{b}{a}}$$

In your case, $$ a=3, b = 3\sqrt{3}, \theta=\pi x$$

I believe it's very easy to continue on from here. Im sure there are many proofs of this online as well, if you need to be convinced that it works.

Answer 5

Your intuition is good, if you realize that the periods are equal so $K=\pi$.

Whenever you add two sinusoids with the same frequency, no matter the phases or amplitudes, you always get another sinusoid with the same frequency.

First I suggest using $\cos(x) = \sin(x + \frac \pi 2)$ to make everything a sum of sines.

$$3\sin(\pi x) + 3\sqrt 3 \cos(\pi x) = A \sin(\pi x + K)$$ $$3\sin(\pi x) + 3\sqrt 3 \sin(\pi x + \frac \pi 2) = A \sin(\pi x + K)$$

A sine wave can be viewed as the height of a rotation circle. Now using the notation $[\text{magnitude} < \text{angle}]$ for polar form, we look at the circles at time $x=0$:

The first circle is at $[3 < 0]$ and the second circle is at $[3 \sqrt{3} < \frac \pi 2]$. The sum is at $[A < K]$. So: $$[3 < 0] + [3 \sqrt{3} < \frac \pi 2] = [A < K]$$

Converting to rectangular form: $$[3\cos(0) , 3\sin(0)] + [3 \sqrt{3} \cos\left( \frac \pi 2\right),3 \sqrt{3} \sin\left( \frac \pi 2\right) ] = [A\cos(K) , A\sin(K)]$$ $$[3, 0] + [0,3 \sqrt{3} ] = [A\cos(K) , A\sin(K)]$$

Solving that, I got $A=6$, $K = \arctan(\sqrt{3}) = \frac \pi 3$.