# sinx + tanx > 2x

• Mar 6th 2008, 08:46 AM
weasley74
sinx + tanx > 2x - help!!
How do I show that sinx + tanx > 2x if 0 < x < pi/2 ?
• Mar 6th 2008, 09:26 AM
colby2152
Quote:

Originally Posted by weasley74
How do I show that sinx + tanx > 2x if 0 < x < pi/2 ?

You want to show 2x is the minimum boundary. Find the derivative of $sin(x)+tan(x)$ and set it equal to zero. Solve for x and find the minimum.
• Mar 6th 2008, 12:00 PM
weasley74
so I get cosx + 1/(cosx)^2 = 0 , and then I get (cosx)^3 + 1 = 0, putting cosx = t, I get t^3 + 1 = 0

and because what must be a minor stroke, i have no idea how to solve that simple equation.
• Mar 6th 2008, 01:28 PM
colby2152
Quote:

Originally Posted by weasley74
so I get cosx + 1/(cosx)^2 = 0 , and then I get (cosx)^3 + 1 = 0, putting cosx = t, I get t^3 + 1 = 0

and because what must be a minor stroke, i have no idea how to solve that simple equation.

Haha, $t=-1$
$x=\pi$

That is not in the range, so we are at a crossroads.
• Mar 6th 2008, 01:38 PM
weasley74
so we didn't really find anything out by doing that?
• Mar 6th 2008, 03:14 PM
iknowone
Both functions are non negative on the specified interval. The derivative of 2x is 2 while the derivative of sin(x) + tan(x) = cos(x)+1/cos(x)^2 which is clearly greater than or equal to 2. Since sin(x)+tan(x) = 2x at x=0 and sin(x)+tan(x) is increasing faster than 2x for every value of x (0,pi/2], sin(x)+tan(x) is greater than or equal to 2x.