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For every $x \in [o,\frac{\pi}{2}]$, $\sin(x)+\cos(x)\geq 1$. How do you prove this rigorously by contradiction?

I understand you start by assuming that $\sin(x)+ \cos(x)<1$ and prove this is a false statement. I can see from drawing the graphs this is false but how do I show it algebraically?

Tk706
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4 Answers4

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Both $\cos(x)$ and $-\sin(x)$ are convex functions on the interval $(\pi/2,\pi)$. Since in the endpoints of such interval se have $\cos(x)-\sin(x)=-1$, it follows that $\cos(x)-\sin(x)\leq -1$ all over $[\pi/2,\pi]$. The last inequality implies that $\sin(x)+\cos(x)$ is a decreasing function over $[\pi/2,\pi]$, so, for any $x\in(\pi/2,\pi]$, $$\sin(x)+\cos(x)<\sin(\pi/2)+\cos(\pi/2)=1.$$

Jack D'Aurizio
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Hint: You may need the formula: $$ \sin x \sin\frac\pi 4+\cos x\cos\frac\pi 4 = \cos\left(x - \frac \pi 4\right) $$

mookid
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  • See the comment under the original question. What the OP is asking is not true. – 5xum Oct 21 '14 at 12:24
  • @5xum: whether it is true or wrong, it allows to have a clear idea what the size of the sum is depending of the interval of $x$ considered. – mookid Oct 21 '14 at 13:42
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suqaring $\sin(x)+\cos(x)<1$ we obtain $\sin(x)^2+\cos(x)^2+2\sin(x)\cos(x)<1$ since $\sin(x)^2+\cos(x)^2=1$ we get $\sin(2x)<0$ which is true since $\frac{\pi}{2}\le x\le \pi$
you can also use that $\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$

Dr. Sonnhard Graubner
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  • I must be doing something wrong. Why is this a contradiction? $\sin{2x}$ is negative on $[\frac{\pi}{2}, \pi]$, isn't it? – layman Oct 21 '14 at 12:16
  • Squaring isn't an order preserving function. – Git Gud Oct 21 '14 at 12:17
  • @MathIsHardNoItsNot You're not doing anything wrong, in fact it's easy to find $x$ in the given interval such that $\sin(x)=-\cos(x)$, yielding $0\ge 1$. – Git Gud Oct 21 '14 at 12:19
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    @GitGud So is this answer wrong then? The only contradiction I can see is if we specifically look at the value $x = \frac{\pi}{2}$, since at that point, you get that the left hand side $=1$, so it is not $< 1$... – layman Oct 21 '14 at 12:20
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    @MathIsHardNoItsNot I was referring to the common in the question, but I'd say the answer is wrong because it's using $\forall a,b(a<b\implies a^2<b^2)$. I didn't read past this. – Git Gud Oct 21 '14 at 12:22
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    since $\sin(x)+\cos(x)\geq 1$ is not true if $\frac{\pi}{2}\le x\le \pi$ – Dr. Sonnhard Graubner Oct 21 '14 at 12:22
  • The only problem is that the proof is done backward: π/2≤ x ≤ π => sin(2x) ≤ 0 => $(sin(x)+cos(x))^2 = sin(2x) +sin(x)^2+cos(x)^2=1 +2sin(2x) ≤ 1$ => |sin(x) +cos(x)| ≤ 1. u ≤ |u| => sin(x) + cos(x) ≤ 1 when π/2≤ x ≤ π – mvggz Oct 21 '14 at 13:45
  • thank you for your comment – Dr. Sonnhard Graubner Oct 21 '14 at 14:10
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We start with the assumption $$\sin x+\cos x<1$$ and rearrange, $$\sin x<1-\cos x$$ square both sides $$\sin ^2 x<1-2\cos x+\cos^2x$$ and manipulate further. $$\sin^2x-\cos^2x<1-2\cos x$$ Since $\sin^2x+\cos^2x=1$ and $\sin^2x+\cos^2x-2\cos^2x=\sin^2x-\cos^2x$, $$1-2\cos^2x<1-2\cos x$$ $$2\cos^2x>2\cos x$$ $$\cos x>1$$ which is false since the codomain of $\cos x$ is $-1\le x\le 1$.

Avi
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