Is it possible to prove $|\sin x| \leqslant |x |$ with only trigonometric identities? (not with well-known calculus or geometry proofs)
Proving $|\sin x| \leqslant |x |$
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calculus
trigonometry
inequality
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3You should start with definition of $sin(x)$ – 2012-11-25
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2Does $\sin^2x+\cos^2x=1$ count as trigonometric identity? – 2012-11-25
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1I guess $|x|>0$... – 2012-11-25
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0Excuse me, I meant $\leqlant$. Yes, this is a trigonometric identitu. – 2012-11-25
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1Two most famous definitions of $\sin x$ make use of either geometric argument or calculus. Thus you have to give a definition that avoids both geometry and calculus in order to suffice your need. – 2012-11-25
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0I assume that sin is allready defined ... – 2012-11-25
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1please use the following definitions $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$ $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$ in this case all finitary trig identities are proved by algebra of rational polynomials in $z = e^{ix}$. – 2012-11-25
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0Cassandrao, I felt the same but I'm not sure. I know what you wrote about finitary trig identities using complex form but I don't know too if this complex form may be used to proof this inequality. Someone told that it can be. I tried to figure out but didn't manage. Or it's a simple proof, or it's false. – 2012-11-25