October 2nd, 2014, 11:06 PM  #1 
Newbie Joined: Apr 2014 From: melbourne Posts: 14 Thanks: 0  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?

October 3rd, 2014, 02:53 AM  #2 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 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.

October 3rd, 2014, 03:05 AM  #3 
Newbie Joined: Apr 2014 From: melbourne Posts: 14 Thanks: 0 
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 ? 
October 3rd, 2014, 03:18 AM  #4 
Newbie Joined: Apr 2014 From: melbourne Posts: 14 Thanks: 0 
Oh yeah, my bad, i am being so silly. Don't worry, i got it already. Thank you so much.

October 3rd, 2014, 03:23 AM  #5 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 
$\displaystyle \frac{\mathbb{e}^{x}+\mathbb{e}^{x}}{2}\cos(x)> \frac{1+x+\frac{x^2}{2}+1x+\frac{x^2}{2}}{2}\cos(x)=\frac{2+x^2}{2}\cos(x)=1+\frac{x^2}{2}\cos(x)>11=0$
Last edited by ZardoZ; October 3rd, 2014 at 03:33 AM. 

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