$\text{ If }\sin{x} + \cos{x} = m\text{, prove:- }\sin^{6}{x} + \cos^{6}{x} = \frac{4 - 3(m^2 -1)^2}{4}\text{, where }m^2 \leq 2$ My question is not exactly what's in the title. The proof is pretty straight forward:-
From $\sin{x} + \cos{x} = m$, we get $3\sin^2{x}\cos^2{x} = \frac{3(m^2 - 1)^2}{4}$.
Consequently, $\sin^{6}{x} + \cos^{6}{x} = 1 - 3\sin^2{x}\cos^2{x} = 1 - \frac{3(m^2 -1)^2}{4}$.
My question is what's the purpose of $m^2 \leq 2$ in the question? I always get extremely scared whenever I see some form of inequality included in a question. It took me about 15 minutes to solve the above problem just because I got scared of the inequality. But as it turns out I did not need it at all. So Why include it in the question at all? Or does my solution have some critical flaw? Any insight will be very helpful.
Additionally, please let me know of any tips for working with inequalities if possible. For example, how to find a possible path to just start because that's the biggest issue with me. I don't seem to be able to start working a problem if there's an inequality, I start panicking for some reason. Not sure if I'm making this last part much clear or not.
 A: The inequality given here is completely useless. Maximum value of $\sin x+\cos x$ is $\sqrt 2$ and minimum is $-\sqrt 2$, so $m^2\leq 2$ is always valid anyway.
Perhaps this was given to mislead you or waste your time, in which case, you fell for the trap!
Another reason this might have been given was to make sure you know that we are only considering real values for $x$ here. This is because, the familiar bounds of $-1\leq \sin x \leq 1$ are only valid for real numbers.
Regarding your perhaps rather irrational fear of inequalities, what is it exactly that concerns you? In many cases inequalities can actually help you, as they help narrow down the available methods. For example, in some cases where the AM-GM inequality is a possible approach, you might be given to work with positive numbers, as AM-GM only works for those.
A: The purpose of this inequality is a sanity check. Since $(\sin(x) + \cos(x))^2 = 1 + \sin(2x) \le 2$ for all real $x$, $m^2 \le 2$ ensures we're only looking at real values of $x$.
The statement is actually true even if we remove the condition $m^2 \le 2$ and allow complex $x$, but I'm guessing complex numbers aren't something you've covered yet.
