I already know:
$$\cos(x)\sin(x)\leq \frac{1}{2}$$
$$\cos(x)+\sin(x)\leq \sqrt{2}$$
How can I use these to prove inequality $\sin(x)^3+\cos(x)^3 \leq 1$? I tried to use binomial expansion, but this doesn't get me anywhere.
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$\begingroup$ Are you sure with your inequalities? $\endgroup$– FakemistakeJan 2, 2020 at 10:48
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$\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$– N. F. TaussigJan 2, 2020 at 10:50
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3 Answers
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$$\sin^3x\le \sin^2x\tag1\label{1}$$ $$\cos^3x\le \cos^2x\tag2\label{2}$$ Add \eqref{1} and \eqref{2} to get the result.
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$\begingroup$ Is that because |cos(x)|≤1? (And same for sine) $\endgroup$ Jan 2, 2020 at 10:48
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2$\begingroup$ You don't need those absolute values, do you ? $\endgroup$– user65203Jan 2, 2020 at 10:49
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2$\begingroup$ @Yves Daoust, No, I don't , but it shows that something stronger is true. $\endgroup$– MartundJan 2, 2020 at 10:50
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This is not an answer but a comment which graph cannot be edited in the comments section.
No need to add something to Martund's smart proof (which can be generalized).
A stronger inequality : $$-1\leq\cos^n(x)+\sin^n(x)\leq 1\qquad n\geq 2$$
answered Jan 2, 2020 at 11:35
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Use $\sin^2(x)+\cos^2(x)=1$ to show $\sin^3(x)+\cos^3(x) = (\sin(x) + \cos(x))(1-\sin(x)\cos(x))$ and then apply the (corrected) inequalities.
