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I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.

Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$

Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.

Getting $r$:

$$\begin{align*} 3^2 + 4^2 &= r^2 \cos(a)^2 + r^2 \sin(a)^2\\ 25 &= r^2 (\cos(a)^2 + \sin(a)^2)\\ 25 &= r^2\\ r &= -5, 5 \end{align*}$$

Getting $a$:

$$\begin{align*} \frac{-r\sin(a)}{ r\cos(a)} &= \frac{3}{4}\\ \tan(a) &= -\frac{3}{4}\\ a &= \arctan(-3/4)\\ a &= -36.87, 143.13\\ \end{align*}$$

The questions:- There are two values for both $r$ and $a$, how should I choose the values to be in the final form?

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What do you mean in the final form? (Btw, +1 for showing your working) – Aryabhata Mar 4 '12 at 3:30
(a) After you get your solution, graph it (and the original function) and see if they agree. (b) Why do you think the problem (or any problem) has only one solution? – GEdgar Mar 4 '12 at 3:31
up vote 1 down vote accepted

The two values of $a$ differ by $180^{\circ}$. Since $\sin(y\pm 180^{\circ}) = -\sin(y)$, you want to pick $r$ and $a$ in such a way that $r\sin(x-a)$ has the same sign as $3\sin x+ 4\cos x$.

When $x=0$, $3\sin x + 4\cos x=4$ is positive; so you want to pick $r$ and $a$ so that $r\sin(-a)$ is positive. This happens when $r=5$ and $a=-36.87^{\circ}$; or when $r=-5$ and $a=143.13$. The two choices work, the other two possibilities do not.

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$$\begin{align*} f(x) &= 3 \sin(x) + 4 \cos(x) \\ f(x) &= 5\left(\frac{3}{5} \sin(x) + \frac{4}{5} \cos(x) \right) \end{align*}$$

Let $\sin(y) = 4/5$ and $\cos(y) =3/5$, for some real number $y$. $$\begin{align*} f(x) &= 5\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\ f(x) &= 5\sin(x+y)&\text{(where }y = \arcsin(4/5)\text{)} \end{align*}$$ Using trigonometric tables, $y = 36.87$. $$f(x) = 5\sin(x + 36.87^{\circ})$$ which can also be written as $$f(x) = 5 \sin (180 - x - 36.87) = -5\sin(x - 144.13).$$ This justifies the 2 values of $a$ and $r$ in $r\sin(x−a)$.

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Thanks for the edit – Tomarinator Mar 4 '12 at 7:25

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