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October 2nd, 2014, 11:06 PM   #1
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Compare sinh(x) and sin(x)

I am having trouble proving the relationship between the hyperbolic function sinh(x) and the trigonometric function sin(x). i believe that sinh(x)>sin(x) for (0,inf) and sinh(x)<sin(x) for (inf,0) and the graph also supports this, but any ideas how to prove this?
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October 3rd, 2014, 02:53 AM   #2
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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
Let $\displaystyle h(x)=\sinh(x)-\sin(x)=\frac{\mathbb{e}^{x}-\mathbb{e}^{-x}}{2}-\sin(x)\Rightarrow h'(x)=\frac{\mathbb{e}^{x}+\mathbb{e}^{-x}}{2}-\cos(x)\geq 0 $ so since h(0)=0 and $h\uparrow$ you have $h(x)>0\Rightarrow \sinh(x)>\sin(x)$ for x>0 and $h(x)<0\Rightarrow \sinh(x)<\sin(x)$ for x<0.
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October 3rd, 2014, 03:05 AM   #3
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thank you for your reply, but one more question though, how do you prove this :
h'(x)=1/2*(e^x+e^-x)-cosx >0 ?
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October 3rd, 2014, 03:18 AM   #4
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Oh yeah, my bad, i am being so silly. Don't worry, i got it already. Thank you so much.
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October 3rd, 2014, 03:23 AM   #5
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$\displaystyle \frac{\mathbb{e}^{x}+\mathbb{e}^{-x}}{2}-\cos(x)> \frac{1+x+\frac{x^2}{2}+1-x+\frac{x^2}{2}}{2}-\cos(x)=\frac{2+x^2}{2}-\cos(x)=1+\frac{x^2}{2}-\cos(x)>1-1=0$

Last edited by ZardoZ; October 3rd, 2014 at 03:33 AM.
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