Solving Sinx Equations for 0 ≤ x ≤ π/2: Proofs and Solutions

[axmls]

Note: This is not homework. This is from a monthly math challenge that my university offers. This particular question is past its due date, and I'm discovering it late. I'm an engineering student who's fascinated by math and wants to give problems from the math department a shot (though I have no idea if I even have the tools to solve some of these, as I've only just finished Calc II).

Homework Statement

(a) Show that there exists exactly one real solution of sinx +sin^2x = 1 for <br /> 0 ≤ x ≤ π/2



(b) Show that for each n ∈ N there exists exactly one real solution of
sin x + (sin^n x) = 1 for 0 ≤ x ≤ π/2

Homework Equations

N/A

The Attempt at a Solution

I've never really had to use proofs in my math classes so far. I assume I'll need the intermediate value theorem.

For part (a), we see that sin(0) + sin^2(0) = 0 &lt; 1 and that
sin(π/2) + sin^2(π/2) = 1 + 1 = 2 &gt; 1. Since f(x) = sinx + sin^2x is continuous for all x, and since f(0) &lt; 1 &lt; f(π/2), there must exist some c ∈ [0, π/2] such that f(c) = 1. This would be good I believe if it weren't for the condition that there is exactly one real solution, which the intermediate value theorem wouldn't be able to let us know. Am I approaching this wrong? Help would be appreciated (as I don't have any background in writing proofs).

Part (b) I feel encounters the same difficulty.

Thanks.

 

[haruspex]

For part (a), we see that sin(0) + sin^2(0) = 0 &lt; 1 and that
sin(π/2) + sin^2(π/2) = 1 + 1 = 2 &gt; 1. Since f(x) = sinx + sin^2x is continuous for all x, and since f(0) &lt; 1 &lt; f(π/2), there must exist some c ∈ [0, π/2] such that f(c) = 1. This would be good I believe if it weren't for the condition that there is exactly one real solution, which the intermediate value theorem wouldn't be able to let us know.

Quite so.
One way: Suppose there are two solutions. That allows you to write down two equations. Can you see how to combine the two equations in an interesting way? Any factorisation?
Another way: solve the quadratic equation. Rule out one of the two solutions based on the range of x.
Either way, you can reduce it the problem of showing that sin(x) is strictly monotonic over that range.

 

[axmls]

So is it only necessary to show that the function is monotonic on the interval [0, π/2]?

For instance, I could say $f(x) = sin(x) + sin^2(x)$
Then $f'(x) = cos(x) + 2sin(x)cos(x) = cos(x) + sin(2x)$

Since $cos(x) ≥ 0$ on $[0, π/2]$ and $sin(2x) ≥ 0$ on $[0, π/2]$, then
$cos(x) + sin(2x) ≥ 0$ on $[0, π/2]$, which means $f$ is strictly increasing on the interval $[0, π/2]$, and therefore there's only one solution to the equation
$sin(x) + sin^2(x) = 1$ (the solution being the one that is shown to exist by the intermediate value theorem).

Would this be a sufficient proof for part (a)?

 

[haruspex]

So is it only necessary to show that the function is monotonic on the interval [0, π/2]?

For instance, I could say $f(x) = sin(x) + sin^2(x)$
Then $f'(x) = cos(x) + 2sin(x)cos(x) = cos(x) + sin(2x)$

Since $cos(x) ≥ 0$ on $[0, π/2]$ and $sin(2x) ≥ 0$ on $[0, π/2]$, then
$cos(x) + sin(2x) ≥ 0$ on $[0, π/2]$, which means $f$ is strictly increasing on the interval $[0, π/2]$, and therefore there's only one solution to the equation
$sin(x) + sin^2(x) = 1$ (the solution being the one that is shown to exist by the intermediate value theorem).

Would this be a sufficient proof for part (a)?

Yes, I think that's fine.
It's often tricky having to guess what properties you're allowed to assume. You're not being given a definition of sin(x) to work from. Knowing the derivative is a relatively advanced fact, so it might be better to avoid it, but it's probably ok here.