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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Compare sinh(x) and sin(x)

• Thread starter phongchaychua
• Start date Oct 3, 2014
• Tags compare sinhx sinx

P

phongchaychua

Joined Apr 2014 14 Posts | 0+ melbourne 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?

ZardoZ

Joined Nov 2010 2K Posts | 137+ Greece, Thessaloniki 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. P

phongchaychua

Joined Apr 2014 14 Posts | 0+ melbourne 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 ? P

phongchaychua

Joined Apr 2014 14 Posts | 0+ melbourne Oh yeah, my bad, i am being so silly. Don't worry, i got it already. Thank you so much.

ZardoZ

Joined Nov 2010 2K Posts | 137+ Greece, Thessaloniki Last edited: Oct 3, 2014 $\displaystyle \frac{\mathbb{e}^{x}+\mathbb{e}^{-x}}{2}-\cos(x)> \frac{1+x+\frac{x^2}{2}+1-x+\frac{x^2}{2}}{2}-\cos(x)=\frac{2+x^2}{2}-\cos(x)=1+\frac{x^2}{2}-\cos(x)>1-1=0$ Login or Register / Reply

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