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I understand that with questions asking you to prove for x on an interval requires induction. The question I have is: prove that for every real number $x \in \left[0,\dfrac{\pi}{2}\right]$, $\sin(x)+\cos(x)\ge1$.

The base case is obvious: 0 satisfies the equation I want to prove.

The inductive case is more confusing. The hypothesis is that there is an n that satisfies the equation. But then what should I do for the n+1 case?

Would I do $n+2\pi$ because $2\pi$ is similar to adding 1?

Théophile
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user2793618
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    I understand that with questions asking you to prove for x on an interval requires induction. Your understanding is wrong. Induction can only prove that a statement is true for positive integers (or, more generally, some countable set of cases). Any interval contains uncountably many reals, so you cannot use induction in that case. – dxiv Aug 26 '18 at 04:06
  • Oh boy. Ok, so how should I proceed with this? – user2793618 Aug 26 '18 at 04:08
  • First of all, provide some context about where the question came from, and what your background is as far as trig, calculus etc. – dxiv Aug 26 '18 at 04:11
  • I've taken AP Calculus BC, but I am new to proofs and logic. – user2793618 Aug 26 '18 at 04:13

2 Answers2

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Hint : $\sin x + \cos x = \sqrt 2 \sin(x + \frac{\pi}4)$

Deepak
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Alt. hint: $\;(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x \ge 1\;$ (why?).

[ EDIT ]   A complete proof would just need to piece together the following:

  • if $a \ge 0$ then $a \ge 1 \iff a^2 \ge 1$

  • if $x \in [0, \pi/2]$ then $\sin x \ge 0$ and $\cos x\ge 0$, so $\sin x + \cos x \ge 0$

  • therefore $\sin x + \cos x \ge 1 \iff (\sin x + \cos x)^2 \ge 1$ for $x \in [0,\pi/2]$

  • $x \in [0, \pi/2] \iff 2x \in [0, \pi]$, and $\sin$ is non-negative on $[0,\pi]$

  • it follows that $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x \ge 1$

dxiv
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  • Because sin(2x) is always going to be >=0. How do I incorporate this into the proof though? I'm specifically confused on how to do it with the "every". – user2793618 Aug 26 '18 at 04:16
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    @user2793618 Yes. More precisely, because $,2x \in [0, \pi],$ and $\sin$ is non-negative on $,[0,\pi],$. – dxiv Aug 26 '18 at 04:17
  • Could you please write out a proof? I'm just really confused on how to format it. – user2793618 Aug 26 '18 at 04:21
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    @user2793618 I edited the steps into the answer above. – dxiv Aug 26 '18 at 04:30
  • A question about induction. So induction questions often ask "for every integer blah blah blah", is induction used in these cases because even though the set of integers is of an infinite number of elements, the "types" (excuse the cs language) are the same? And in this case with the interval, the types are not the same (integers and decimals)? – user2793618 Aug 26 '18 at 04:34
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    @user2793618 That question goes a lot deeper, in fact. But for a quick intuitive argument, consider that induction is a way to cover a set of cases step-by-step, with each step relying on the previous ones. This works for integers, because every integer $n \in \Bbb Z$ has a "next" integer $n+1$, so you can "step through" all of them. But induction does not (and cannot possibly) work for reals, since there is no "next" real to $x \in \Bbb R$, and no way to (meaningfully) define such a "next". – dxiv Aug 26 '18 at 04:42
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    Oh I see. It's because induction relies on the n+1. So induction only works with integers? – user2793618 Aug 26 '18 at 04:43
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    @user2793618 Short answer, yes. Long answer depends on how exactly you define "induction" itself, but it will never work for an uncountable set like an interval of reals. – dxiv Aug 26 '18 at 04:46
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    Great, thanks!! – user2793618 Aug 26 '18 at 04:48