# How does teacher get first step?

Below are the steps the teacher took to solve:

$y = \sqrt{3}\sin x + \cos x$ find min and max on $[0, 2\pi)$

Step 1: = $2\sin(x + \pi/6))$

How does the teacher get this first step?

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I'm posting this answer in response to the comment thread under picakhu's answer; writing comments was getting a bit tedious. The answer has the same content as Isaac's, but is explained a little differently, which might be useful for the OP.

The general problem, of which this is a special case, is:

Given an expression of the form $a \sin x + b \cos x$, for some numbers $a$ and $b$, to rewrite it in the form $c \sin (x + \theta)$, for some appropriate choice of $c$ and $\theta$ (which will be related to $a$ and $b$ in some way, of course --- and we have to find out what that way is!).

Suppose first that $a^2 + b^2 = 1$. Then trigonometry (especially the point of view of the unit circle) tells us that there is an angle $\theta$ such that $a = \cos \theta$ and $b = \sin \theta$.

Thus we can write $a \sin x + b \cos x = \cos \theta \sin x + \sin \theta \cos x,$ and via the addition theorem for $\sin$, we recognize this as being $\sin(x + \theta).$

Now in most examples, it may not be that $a^2 + b^2 = 1$. But it equals something (!), so let's call that something $c^2$. Then we see that $(a/c)^2 + (b/c)^2 = 1$, so we can choose $\theta$ so that $a/c = \cos \theta$ and $b/c = \sin \theta$. Then as above we find that $(a/c)\sin x + (b/c) \cos x = \sin (x + \theta),$ and so (finishing finally) we have $$a \sin x + b \cos x = c \sin (x + \theta).$$

In the OP's example, $a = \sqrt{3}$ and $b = 1$, so $c = 2$, and we are led to the given answer.

Note that in practice this whole process will be easiest when $a/c$ and $b/c$ are easily recognized trig function special values, like $1/2$ or $\sqrt{2}/2$ or $\sqrt{3}/2$. If they are just somewhat random numbers, then you won't be able to figure out the correct choice of $\theta$ without using a calculator to compute $\theta$.

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Your answer is easier for me to follow. –  O.O Jan 16 '11 at 17:01

picakhu's answer is the simplest way to see how it works having already arrived at $y=2\sin(x+\frac{\pi}{6})$ (use the identity there to expand this form). In general, given $a\sin x+b\cos x$ (let's say for $a,b>0$), it is possible to arrive at a similar equivalent form: \begin{align} a\sin x+b\cos x &=a\left(\sin x+\frac{b}{a}\cos x\right) \\ &=a\left(\sin x+\tan\left(\arctan\frac{b}{a}\right)\cos x\right) \\ &=a\left(\sin x+\frac{\sin\left(\arctan\frac{b}{a}\right)}{\cos\left(\arctan\frac{b}{a}\right)}\cos x\right) \\ &=\frac{a}{\cos\left(\arctan\frac{b}{a}\right)}\left(\sin x\cos\left(\arctan\frac{b}{a}\right)+\sin\left(\arctan\frac{b}{a}\right)\cos x\right) \\ &=\sqrt{a^2+b^2}\sin\left(x+\arctan\frac{b}{a}\right). \end{align}

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Instead of telling you how your teacher got that, try doing the opposite and expand out $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$
@subt13: Dear subt13, You have to choose some particular $a$ and $b$. Do you see any similarities between $\sin(a+b)$ and $\sin(x + \pi/6).$ –  Matt E Jan 16 '11 at 16:33
@subt13: $\frac{\pi}{6}$ is not just a random guess at what value to try—the $\sqrt{3}$ suggests something like $\frac{\pi}{6}$ or $\frac{\pi}{3}$, but a value can be found explicitly as in my answer. –  Isaac Jan 16 '11 at 16:44