$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
Suffices to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can't find how to continue from here.
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2$|a+b|\leq \sqrt{a^2+b^2}$ is not true, try $a=3, b=4$ – gammatester Jun 21 '16 at 10:44
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@gammatester Find $x : \sin x = \cos x = 1$... – Abstraction Jun 21 '16 at 10:44
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@Abstraction: That is irrelevant. You cannot prove the inequality, and therefore OP's reasoning cannot be OK. – gammatester Jun 21 '16 at 10:47
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@gammatester Oh, sorry, missed that point. Reasoning is indeed faulty. – Abstraction Jun 21 '16 at 10:49
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@gbox, use https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity – lab bhattacharjee Jun 21 '16 at 10:51
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In fact, $|a\sin x+b \cos x|\leq |a+b|$ is also false : you can take $a=1=-b$ and $x=0$, for example. But $|a\sin x+b \cos x|\leq |a|+|b|$ is true (but since $|a|+|b|\geq \sqrt{a^2+b^2}$, this isn't really helpful here). – Arnaud D. Jun 21 '16 at 11:01
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See this answer, and/or this entry at Trigonography.com. – Blue Nov 04 '17 at 04:36
6 Answers
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Use Cauchy-Schwarz inequality : \begin{align}|a\sin x+b \cos x| = |(a,b)\cdot (\sin x,\cos x)| &\leq \sqrt{a^2+b^2}\sqrt{\sin^2 x+\cos^2 x}=\sqrt{a^2+b^2}.\end{align}
Daniel Fischer
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Arnaud D.
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$$|a \sin x + b \cos x|=\big|\sqrt{a^2+b^2}\left(\frac a{\sqrt{a^2+b^2}}\sin x+\frac b{\sqrt{a^2+b^2}}\cos x\right)\big|=$$ $$=\sqrt{a^2+b^2}|\left(\sin(x+\phi)\right)|\le\sqrt{a^2+b^2}$$
Roman83
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3Ok. Let $\frac a{\sqrt{a^2+b^2}}= \sin \phi, \frac b{\sqrt{a^2+b^2}}=\cos \phi$. Really, $$\sin^2 \phi+\cos^2 \phi=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$$ and $|\sin \phi|=\frac a{\sqrt{a^2+b^2}}\le1$ Then $$sin x \sin \phi+\cos x \cos \phi=\sin(x+\phi)$$ – Roman83 Jun 21 '16 at 11:08
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Find the extrema of
$$a\cos(x)+b\sin(x).$$
By canceling the derivative,
$$-a\sin(z)+b\cos(z)=0$$ or$$\tan(z)=\frac ba.$$
Then, with
$$\cos(z)=\pm\frac1{\sqrt{\tan^2(z)+1}}=\pm\frac a{\sqrt{a^2+b^2}},\\\sin(z)=\pm\frac{\tan(z)}{\sqrt{\tan^2(z)+1}}=\pm\frac b{\sqrt{a^2+b^2}},$$
you obtain
$$a\cos(z)+b\sin(z)=\pm\sqrt{a^2+b^2}.$$
Then for all $x$,
$$-\sqrt{a^2+b^2}\le a\cos(x)+b\sin(x)\le\sqrt{a^2+b^2}.$$
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$$a\sin{x}+b\cos{x}=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin{x}+\frac{b}{\sqrt{a^2+b^2}}\cos{x}$$ now let $$\sin{\alpha}=\frac{b}{\sqrt{a^2+b^2}}$$ then $$\cos{\alpha}=\frac{a}{\sqrt{a^2+b^2}}$$ thus $$a\cos{x}+b\sin{x}=\sqrt{a^2+b^2}\sin({\alpha+x})<\sqrt{a^2+b^2}$$
avz2611
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Take two vectors: $v_1 = (a,b)$ and $v_2 = (\sin x, \cos x)$. Their scalar product is $(v_1, v_2) = |v_1||v_2|cos(\phi) = a \sin x + b \cos x$ where $\phi$ is the angle between $v_1$ and $v_2$, but $|v_1| = \sqrt{a^2 + b^2}, |v_2| = 1$.
Abstraction
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Don't see any reason for a down vote. Canceling it by a +1. Nice answer – Shailesh Jun 21 '16 at 11:36
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If we square the inequality we get $$a^2\sin^2 x+b^2\cos^2 x+2ab\sin x\cos x\leq a^2+b^2\\2ab\sin x\cos x\leq a^2(1-\sin^2 x) +b^2(1-\cos^2 x) =a^2\cos^2x+b^2\sin ^2 x\\0\leq a^2\cos^2x-2ab\sin x \cos x+b^2\cos^2 x=(a\cos x-b\sin x) ^2$$ The last equation is always true, and we can square inequality because both sides are positive.
kingW3
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