find minimum of given function today my relative asked a  problem,which had strange solution and i am curious, how this solution  is get from such kind of  equations. let say function has form
$f(x)=a\sin(x)+b\cos(x)$
we should find it's minimum,we have not any constraints or something like this,as i know to find minimum,we should find point where it  reaches minimum and then put this point into first equation,so in our case we have
$f'(x)=a\cos(x)-b\sin(x)$
 or when   we set this  to zero and  also  convert in tangent form ,we get
$\tan(x)=a/b\\;\text{or}\;x=tan^{-1} (a/b)$
now if we put this into first equation,it would be difficult without  calculator to calculate minimum,let say $a=3$ and $b=2$, but my relative told me there exist  such kind of formula that minimum is directly  $\sqrt{a^2+b^2}$, in our case  $\sqrt{13}$, is it right? first of all  i think that we can get  value  $3$, if  $\alpha=0$; 
please help me
 A: Let's take $\phi$ such that $\cos \phi=\frac{a}{\sqrt{a^2+b^2}}$, $\sin \phi=\frac{b}{\sqrt{a^2+b^2}}$, then our function writes
$$\sqrt{a^2+b^2}\sin x \cos\phi+\sqrt{a^2+b^2}\sin\phi\cos x=\sqrt{a^2+b^2}\sin(x+\phi).$$ Now it's easy to minimize it.
A: It appears that the main confusion here is how to find
$$\sin(\tan^{-1}(a/b))\quad \text{or}\quad \cos(\tan^{-1}(a/b))$$
Well, we have $\tan(\theta) = \frac{a}{b}$, for some $\theta$.  So, let's create a right triangle like that, where the opposite side has length $a$, and the adjacent side has length $b$.

We know that $\sin(\theta) = \frac{\text{opp}}{\text{hyp}}$, so then:
$$\sin(\theta) = \frac{a}{\sqrt{a^2 + b^2}}$$
We also know that $\tan^{-1}(a/b)=\theta$, so then:
$$\sin(\tan^{-1}(a/b)) = \frac{a}{\sqrt{a^2+b^2}}$$
The procedure for cosine is similar.
A: Once you have $x_0 = \tan^{-1}(a/b)$ we can throw that back into $\sin(x)$ and $\cos(x)$ without trouble.
Notice that if $x_0$ was the angle value that gave you $a/b$ after applying tangent, then we can think of $x_0$ as an angle of a right triangle with opposite side '$a$' and adjacent side '$b$'.  This would give us the hypotenuse $\sqrt{a^2 + b^2}$.
Thus applying this function to our particular $x_0$ we find that $$f(x_0) = a \cdot \frac{a}{\sqrt{a^2+b^2}} + b \cdot \frac{b}{\sqrt{a^2+b^2}}$$ which after a quick calculation we find $$f(x_0) = \sqrt{a^2+b^2}.$$
A: If you want to know how to derive that formula, here it is: We have $f(x)=a \sin x +b \cos x$ . Hmm, this looks strangely familiar to the identity $r\sin (x+t)= r(\sin x \cos t + \sin t \cos x)$ , and if our function was of the form  $f(x)=r\sin (x+t)$ then the maximum would be easy to find. It would be $r$ which is the amplitude, right? So let's see if we can get our function into a form of $r\sin (x+t)$ . We multiply our $f(x)$  by a number $r$, so we have $rf(x)=ra \sin x +rb \cos x$ . Because we want to have $r f(x) =\sin x \cos t +\cos x \sin t$ .  We want $ra=\cos t$ and $rb=\sin t$ . We know that $\sin^2 t +\cos^2 t =1$ , so $r^2 a^2+r^2 b^2=1$ $\to$ $r= \large \frac {1}{a^2+b^2}$ . Therefore we have $\large \frac {1}{a^2+b^2} f(x)= \sin (x+t)$ $\to$ $f(x)=(a^2+b^2) \sin(x+t)$ . So the maximum of $f(x)$ $\,$ is $a^2+b^2$ . However, there is one more thing I didn't explain; the motivation behind multiplying everything by the constant $r$ . This is because your example was $3 \sin x +2\cos x$ . If we tried to use the method directly before multiplying by $r$ we would get $\cos t=3$ and $\sin t =2$ , which is impossible because the max of both of these functions is $1$ . Even if $a, b$ were fractions, such as  $\frac {3}{10}$ and  $\frac 56$ ,  they still needed to satisfy $a^2+b^2=1$ . If anything is unclear let me know. Hope this helps.
