Calculus Calculus Math Forum

 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)
 October 3rd, 2014, 02:53 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 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: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus $\displaystyle \frac{\mathbb{e}^{x}+\mathbb{e}^{-x}}{2}-\cos(x)> \frac{1+x+\frac{x^2}{2}+1-x+\frac{x^2}{2}}{2}-\cos(x)=\frac{2+x^2}{2}-\cos(x)=1+\frac{x^2}{2}-\cos(x)>1-1=0$ Last edited by ZardoZ; October 3rd, 2014 at 03:33 AM. Tags compare, sinhx, sinx ,

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relation connecting sinx and sinhx

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