$$\begin{align*} \sin (a) + \cos(a) &\leq \sqrt{2}\\ (\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\ \sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2} \end{align*}$$ Am I doing it right? I need help.
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This question as written is completely unreadable. For some basic information about writing math at this site see e.g. here, here, here and here. – AlexR May 29 '15 at 22:17
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4Have I interpreted the question correctly? – Zev Chonoles May 29 '15 at 22:17
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@ZevChonoles Wow, that's what I call a wild guess :'D – AlexR May 29 '15 at 22:18
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2You're doing it sort of right, but backwards. You have $\sin^2a+\cos^2a = 1$, and $\sin 2a \leq 1$, so $\sin^2a+\cos^2a+\sin2a \leq 2$. Hence $\sin^2a+2\sin a\cos a+\cos^2a \leq 2$. Hence $(\sin a+\cos a)^2 \leq 2$. Hence $\sin a + \cos a \leq \sqrt{2}$. – Brian Tung May 29 '15 at 22:22
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It looks like you want to show $|\sin(a)+\cos(a)|\le \sqrt{2}$. – i. m. soloveichik May 29 '15 at 22:22
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Incidentally, upvoting this because the question seems reasonable, and the asker provided context; it was probably downvoted because of its ungainly appearance? – Brian Tung May 29 '15 at 22:23
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@i.m.soloveichik: That also follows, but I figured the OP can modify the last line of reasoning suitably. :-) – Brian Tung May 29 '15 at 22:23
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@BrianTung It was most likely downvoted because it was completely unreadable in the OPs version. (See the revision history) – AlexR May 29 '15 at 22:26
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thanks brian but can you post it as an answer plz to follow steps – user244454 May 29 '15 at 22:26
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@AlexR: That's sort of what I meant by "ungainly appearance." :-) – Brian Tung May 29 '15 at 22:30
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1By the way, @ZevChonoles, that is an inspired piece of mind-reading and editing. I want to upvote that edit. – Brian Tung May 29 '15 at 22:33
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@user244454: egreg's answer already has that covered. – Brian Tung May 29 '15 at 22:34
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1@Brian: Haha thanks :) – Zev Chonoles May 29 '15 at 22:44
2 Answers
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Alternative path: $$ \sin a+\cos a= \sqrt{2}\left(\sin a\cos\frac{\pi}{4}+\cos a\sin\frac{\pi}{4}\right)= \sqrt{2}\sin\left(a+\frac{\pi}{4}\right)\le\sqrt{2} $$ (and also $\ge-\sqrt{2}$, of course).
However your reasoning is basically correct; only you need to do it backwards: $$ \sin^2a+2\sin a\cos a+\cos^2a\le2 $$ because $\sin^2a+\cos^2a=1$ and $2\sin a\cos a=\sin 2a\le 1$; therefore $$ (\sin a+\cos a)^2\le 2 $$ and so $$ \sin a+\cos a\le \sqrt{2} $$
egreg
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If @Zev Chonoles guessed right, you can prove this by computing $$\max_{x\in [0,2\pi]} \sin x + \cos x$$ The critical points are precisely those where $\cos x - \sin x = 0$ (derivative). These are $x = \frac\pi4$ and $x = \frac{5\pi}4$. Evaluating at those and the boundary points gives $$\sin \frac\pi4 + \cos \frac\pi4 = \frac1{\sqrt2} + \frac1{\sqrt2} = \sqrt2 \\ \sin \frac{4\pi}4 + \cos \frac{5\pi}4 = -\frac1{\sqrt2} - \frac1{\sqrt 2} = -\sqrt2 \\ \sin 0 + \cos 0 = 0 + 1 = 1\\ \sin 2\pi + \cos 2\pi = 0 + 1 = 1$$ Thus the maximum is $\max(\sqrt 2, -\sqrt 2, 1, 1) = \sqrt2$, wich proves $$\sin x + \cos x \le \sqrt2 \qquad\forall\ x\in[0,2\pi]$$ Now since the function is $2\pi$-periodic, we can conclude $$\sin x + \cos x \le \sqrt2 \qquad \forall\ x\in\mathbb R$$
AlexR
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