# Unconventional way to solve trig question

I was working on a trig question and got stuck, but then I noticed a possible way to solve the problem. However, this way seemed to be slightly unconventional and possibly not what the book was looking for.

The question was: "Find $k$ and $b$ when $\sqrt3 \sin x + \cos x=k\sin(x+b)$" The first thing I did was to expand $k\sin(x+b)$ to $k\sin x\cos b+k\cos x\sin b$. Here is where I got stuck. I tried several different things, but I hit dead ends. Then I tried to equate coefficients and got $k\cos b=\sqrt3$ and $k\sin b=1$ which simplified to: $\cos b=\displaystyle{\sqrt3 \over k}$ and $\sin b=\displaystyle{1\over k}$. The answer from there is fairly simple to get which is $k=2$ and $b=\displaystyle{\pi\over6}$.

However this method seems rather dubious and so I was wondering if someone new a better way of solving the problem using more rigorous mathematical methods.

• What makes it seem like a guessing game? Also, I believe you missed a solution. – Mike Dec 18 '11 at 1:54
• $k^2=4$ has two solutions. – Henry Dec 18 '11 at 1:54
• Good point. I just thought that using that method might not work all the, but I guess it might. – E.O. Dec 18 '11 at 1:56
• Equating coefficients works here, but can be avoided by looking at the cases $x=0$ and $x=\pi/2$ and checking the final result. – Henry Dec 18 '11 at 1:56

It's not at all dubious. The functions sin and cos are linearly independent and so $a_1 \sin x + b_1 \cos x = a_2 \sin x + b_2 \cos x$ for all $x$ iff $a_1=a_2$ and $b_1=b_2$.

To see that sin and cos are linearly independent, take $a \sin x + b \cos x = 0$ and set $x=0$ to get $b=0$ and $x=\pi/2$ to get $a=0$.

• This appears to be the only answer that explains why the OP's method is valid. +1 – GeoffDS Dec 18 '11 at 2:46
• @Graphth, Henry's comment was also about this I think. – lhf Dec 18 '11 at 4:01
• what about the case where $x=\pi /3$ – Ramana Venkata Dec 18 '11 at 5:17

I'm not sure how it's perceived as dubious. If you have $k \cos b = \sqrt{3}$ and $k \sin b = 1$, then you must $\frac{k \sin b}{k \cos b} = \tan b = \frac{1}{\sqrt{3}}$, which gets you $b = \frac{\pi}{6}$ or $\frac{7\pi}{6}$. From there it immediately follows that $k = 2$ or $-2$, respectively.

If you have $k\cos b=\sqrt{3}$ and $k\sin b=1$, you should also remember that $k^2\cos^2 b+ k^2\sin^2 b= k^2$. That means $\sqrt{3}^2 + 1^2 = k^2$, so $k=\pm 2$. If you take $k=2$, you can find a suitable value of $b$, and with $k=-2$, you can find another.

Suppose that $a\sin(x)+b\cos(x)=c\sin(x)+d\cos(x)$ holds for all $x$ in some neighborhood of $x_0$. Then we have the following equation and its derivative: \begin{align} 0&=(a-c)\sin(x_0)+(b-d)\cos(x_0)\tag{1}\\ 0&=(a-c)\cos(x_0)-(b-d)\sin(x_0)\tag{2} \end{align} Adding the squares of $(1)$ and $(2)$ gives $$0=(a-c)^2+(b-d)^2\tag{3}$$ Thus, $a=c$ and $b=d$.

So if we have \begin{align} \sqrt{3}\sin(x)+\cos(x) &=k\sin(x+b)\\ &=k\cos(b)\sin(x)+k\sin(b)\cos(x)\tag{4} \end{align} for all $x$ in some neighborhood of $x_0$, then we must have $$k\cos(b)=\sqrt{3}\quad\text{ and }\quad k\sin(b)=1\tag{5}$$ and $(5)$ was solved above.

Here is what I come up in my mind, maybe this is an unconventional way too: Divide both sides by $2$, we get $$\frac{\sqrt3}{2} \sin x +\frac{1}{2}\cos x=\frac{k}{2}\sin(x+b).$$ Note that $\displaystyle\sin(\frac{\pi}{6})=\frac{1}{2}$ and $\displaystyle\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, the left hand side is given by $$\cos(\frac{\pi}{6})\sin x +\sin(\frac{\pi}{6})\cos x=\sin(x+\frac{\pi}{6}),$$ where the last equality follows from compound angle forumula. Combining these, we have $$\sin(x+\frac{\pi}{6})=\frac{k}{2}\sin(x+b).$$

Here is a standard trick to deal with expressions of the type $A \sin(x) + \cos(x)$:

Let $u$ be so that $\cot(u) =A$.

Then, after bringing everything to the same denominator, your expression becomes

$$A \sin(x) + \cos(x)= A\frac{\cos(u) \sin(x)+\sin(u)\cos(x)}{ \sin(u)} =\frac{ \sin(x+u)}{\sin(u) }$$

If you look over the other answers, this is exactly what they did, in a non-explicit way.

Remark If instead you use $\tan(v)=A$, you end up with

$$A \sin(x) + \cos(x) = \frac{\cos(x-v)}{\cos(v)} \,.$$

Same substitutions also solve the expressions of the type $\sin(x) + B\cos(x)$.

In general, \begin{align*} \alpha \sin(x) + \beta \cos(x) &= \sqrt{\alpha^2+\beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}\sin(x) + \frac{\beta}{\sqrt{\alpha^2+\beta^2}}\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\sin(x+y) \end{align*} where $y$ is the angle whose cosine is $\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}}$ and whose sine is $\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}$. In your problem, $\alpha = \sqrt{3}, \beta = 1$, which readily gives $k = 2$, $b = \frac{\pi}{6}$. As others have pointed out, this misses one of the two solutions.
If $\sqrt{\alpha^2+\beta^2}$ (which conventionally means the positive square root) is replaced throughout by $({\alpha^2+\beta^2})^{1/2}$, we get the other solution $k = -2, b = \frac{7\pi}{6}$ upon choosing $({\alpha^2+\beta^2})^{1/2} = -2$.