# Homework Help: Range of the function

1. Aug 30, 2013

### Saitama

1. The problem statement, all variables and given/known data
$\forall x \in R$, find the range of function, $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$; $\alpha \in [0,\pi]$.

2. Relevant equations

3. The attempt at a solution
Honestly, I cannot think of how to approach this problem. I don't see how would I simplify the given expression. I require a few hints to start with.

Any help is appreciated. Thanks!

2. Aug 30, 2013

### drewfstr314

My advice would be to break it down and look at each part individually. Another strategy that probably won't work here is that the range of f(x) is the same as the domain of f^-1(x) [f-inverse]. Since domain is easier to find, that might work.

3. Aug 30, 2013

### Saitama

I did think of finding the inverse but then I couldn't find a way to do that.

4. Aug 30, 2013

### drewfstr314

$f(x) = \cos x (\sin x + \sqrt{\sin^2 x + \sin^2 \alpha})$

Well, to find the inverse, switch x and y, and solve for y = f^-1(x):

$x = \cos y (\sin y + \sqrt{\sin^2 y + \sin^2 \alpha})$

I think the first step is to get rid of the square root by isolating it:

$x\sec y - \sin y = \sqrt{\sin^2 y + \sin^2 \alpha}$

Can you solve for y now?

Squaring each side,

$x^2\sec^2 y - 2x\sec y \sin y + \sin^2 y = \sin^2 y + \sin^2 \alpha$

sec y sin y = sin y * 1/cos y = tan y, so

$x^2\sec^2 y - 2x\tan y = \sin^2 \alpha$

Then, there's a trig identity 1 + tan^2 x = sec^2 x, so

$x^2(1 + \tan^2 y) - 2x\tan y - \sin^2 \alpha = 0$

Distributing and rearranging,

$x^2\tan^2 y - 2x\tan y + x^2 - \sin^2 \alpha = 0$

You might recognize this as a quadratic equation, not in x, but in tan y. We can use the quadratic formula:

$\tan y = \frac{2x \pm \sqrt{4x^2 - 4(x^2)(x^2 - \sin^2 \alpha)}}{2(x^2)}$

$\tan y = \frac{2x \pm \sqrt{4x^2 - 4x^4 + 4x^2\sin^2 \alpha}}{2x^2}$

Factoring out an 4x^2 from the square root,

$\tan y = \frac{2x \pm \sqrt{x^2(1 - x^2 + \sin^2 \alpha}}{2x^2}$

Then, $\sqrt{a^2b} = a\sqrt{b}$, so

$\tan y = \frac{2x \pm 2x\sqrt{1+\sin^2 \alpha - x^2}}{2x^2}$

Simplifying this,

$\tan y = \frac{1 \pm \sqrt{1+\sin^2 \alpha - x^2}}{x}$

Then, (tan y = tan(y + pi) = tan(y + 2pi) = ...),

$y = \arctan\left(\frac{1 \pm \sqrt{1+\sin^2 \alpha - x^2}}{x} \right) + k\pi$

where arctan(x) is the principal value of the inverse-tangent.

And we can finally see what the domain will be.

Domain of arctan(x): all reals

Domain of inside parts:
- The value under the square root has to be non-negative:
$1+\sin^2 \alpha - x^2 \geq 0$
$x^2 \leq 1+\sin^2 \alpha$

For these kinds of inequalities (x^2 <= constant), the interval is always [-sqrt(constant), sqrt(constant)], so

$-\sqrt{1 + \sin^2 \alpha} \leq x \leq \sqrt{1+\sin^2 \alpha}$

- The denominator cannot be zero.
$x \neq 0$

So the domain of y is simply where both of these is true. But we need to figure out what sin^2(α) is. The bounds are completely irrelevant, since sin^2(x) has range [0, 1]. So the values under the square roots is 1 + [0, 1] = [1, 2]. The smallest negative value in the range is when sin^2(α) = 1, and -sqrt(1+1) = -sqrt(2). The largest positive value in the range is also when sin^2(α) = 1 and sqrt(1+1) = sqrt(2). But we can't include zero. So the range is $[-\sqrt{2}, 0) \cup (0, \sqrt{2}]$.

I've looked at WolframAlpha on this problem, but the graph does go through zero (which is easy to see by setting f(x) = 0. cos x is zero at pi/2 + 2pi*k, so it goes through zero infinitely many times). The actual range is [-sqrt(2), sqrt(2)] with zero, but I don't know why there is an x in the denominator leaving it out of the domain.

5. Aug 31, 2013

### Saitama

According to the given answer, range is $[ -\sqrt{1+\sin^2\alpha},\sqrt{1+\sin^2\alpha}]$ but the inverse include x in denominator so the range should not be equal to zero.

Anyways, thanks a lot drewfstr314! :)

6. Aug 31, 2013

### BruceW

you can see it in the first line of the calculation of the inverse.
$xsec(y)-sin(y)= \sqrt{sin^2(y)+sin^2( \alpha )}$
Now, it looks like x=0 is not allowed, since we want to have an arbitrary alpha. But there is also the possibility that sec(y) goes to infinity. Or, in a more mathematically rigorous way, when we have cos(y)=0 then we are not allowed to divide by cos(y). So, really the 'inverse' is not allowed for certain values.

edit: blech, I'm talking rubbish. It would still be possible to have $x=0$ and $\alpha=0$ and this is still allowed. But anyway, I think that somewhere in the calculation there will be a step similar to this, which effectively does something that is not allowed for certain x values.

edit2: yeah ok, I found it. it is when the quadratic formula is used to give:
$\tan y = \frac{2x \pm \sqrt{4x^2 - 4(x^2)(x^2 - \sin^2 \alpha)}}{2(x^2)}$
But this 'quadratic formula' already assumes that x is not zero. i.e. it does not faithfully 'replicate' the true function at x=0.

Last edited: Aug 31, 2013
7. Aug 31, 2013

### pasmith

Let $y = \cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$. Then we can re-arrange this as in drewfstr314's post (without first interchanging x and y) to get
$$y^2 \tan^2 x - 2y\tan x + y^2 - \sin^2\alpha = 0$$
This is a quadratic in $\tan x$. Given $y$, there will exist $x$ such that $f(x) = y$ if and only if that quadratic has real roots (because for any $C \in \mathbb{R}$ there exists $x \in \mathbb{R}$ such that $\tan x = C$). This will be the case if and only if the discriminant is non-negative. This in turn gives you an inequality which $y^2$ must satisfy, and the range of $f$ is exactly the set of $y$ which satisfy that inequality.

8. Sep 7, 2013

### juantheron

Let $$y = f(x) = \cos x \left(\sin x + \sqrt{\sin^2 x+\sin^2 \alpha}\right)$$

$$y = \sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}$$

Now Using Cauchy-Schwartz Inequality, We Get

$$\left(\sin^2 x+\cos^2 x\right).\left(\cos^2 x+ \left(\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2\right)\geq \left(\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2$$

$$\left(1+\sin^2 \alpha\right) \geq \left(\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2$$

OR $$\left(\sin x.\cos x+\cos x.\sqrt{\sin^2 x+\sin^2 \alpha}\right)^2 \leq \left(\sqrt{1+\sin^2 \alpha}\right)^2$$

So $$y^2 \leq \left(\sqrt{1+\sin^2 \alpha}\right)^2$$

So $$-\sqrt{1+\sin^2 \alpha} \leq y \leq +\sqrt{1+\sin^2 \alpha}$$

So $$y \in \left[-\sqrt{1+\sin^2 \alpha},+\sqrt{1+\sin^2 \alpha}\right]$$

9. Sep 7, 2013

### Saitama

Excellent solution juantheron!!!!

Never thought I could apply Cauchy-Schwarz here. Thanks a lot!