# Finding the minimum value

• Nov 28th 2012, 07:02 PM
gfbrd
Finding the minimum value
Hey there I need some help with this problem

Find the minimum value of
absolute value of sinx + cosx + tanx + cotx + secx + cscx

I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks.
• Nov 29th 2012, 09:34 PM
x3bnm
Re: Finding the minimum value
Quote:

Originally Posted by gfbrd
Hey there I need some help with this problem

Find the minimum value of
absolute value of sinx + cosx + tanx + cotx + secx + cscx

I've been trying to combine them together to make it more simpler but I been have trouble doing that, so Im hoping if someone can guide me through this problem so I can understand how to do this thanks.

The derivative of:

$\displaystyle f(x) = \sin(x)+ \cos(x) + \tan(x) + \cot(x) + \sec(x) + \csc(x)$

is:

$\displaystyle \frac{df}{dx} = \cos(x)-\sin(x)+\tan^{2}(x)-\cot^{^2}(x)+\sec(x) \tan(x)-\csc(x) \cot(x)$

The $\displaystyle \frac{df}{dx}$ above can be written as(for $\displaystyle y > 0$):

\displaystyle \begin{align*}=& \cos(x) - \sin(x) + \frac{\sin^{2}(x)}{\cos^{2}(x)} -\frac{\cos^{2}(x)}{\sin^{2}(x)} + \left(\frac{\sin(x)}{\cos(x)}\right)\left(\frac{1} {\cos(x)}\right) - \left( \frac{\cos(x)}{\sin(x)}\right)\left(\frac{1}{\sin( x)}\right) \\ \\ =& \cos(x) - \sin(x) + \frac{\sin^{2}(x)}{\cos^{2}(x)} -\frac{\cos^{2}(x)}{\sin^{2}(x)} + \frac{\sin(x)}{\cos^{2}(x)} - \frac{\cos(x)}{\sin^{2}(x)} \\ \\ =& \cos(x) - \sin(x) + \frac{\sin^{4}(x) - \cos^{4}(x)}{\sin^{2}(x) \cos^{2}(x)} + \frac{\sin^{3}(x) - \cos^{3}(x)}{\sin^{2}(x) \cos^{2}(x)} \\ \\ =& \frac{\sin^{2}(x) \cos^{2}(x)[\cos(x) - \sin(x) + \sin^{2}(x) - \cos^{2}(x) + \sin(x) - \cos(x)]}{\sin^{2}(x) \cos^{2}(x)}\\ \\ =& \cos(x) - \sin(x) + \sin^{2}(x) - \cos^{2}(x) + \sin(x) - \cos(x)\\ \\ =& -\cos(2x)\end{align*}

Now to find absolute min set $\displaystyle \frac{df}{dx} = 0$ and solve for $\displaystyle x$:

\displaystyle \begin{align*}-\cos(2x) =& 0 \\ 2x =& \cdots, -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \cdots \\ x =& \cdots, -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, \cdots\end{align*}

same for $\displaystyle y < 0, \frac{df}{dx}$ will be:

$\displaystyle \frac{df}{dx} = \cos(2x) = 0$

So plugin the value of $\displaystyle x$ one by one and test which one's the smallest. You can see that there's a pattern.

Absolute min value is: $\displaystyle f(\frac{3\pi}{4}) = 2$

Hope it helps.
• Nov 29th 2012, 10:14 PM
gfbrd
Re: Finding the minimum value
Thanks a lot for your help
• Nov 30th 2012, 01:00 AM
BobP
Re: Finding the minimum value
f(2.65345) = -1.82843 (5dp).
• Nov 30th 2012, 03:23 AM
x3bnm
Re: Finding the minimum value
Quote:

Originally Posted by BobP
f(2.65345) = -1.82843 (5dp).

BobP you're right. Just curious can you kindly tell me how you got that mathematically?
• Nov 30th 2012, 03:47 AM
BobP
Re: Finding the minimum value
Newton-Raphson on the derivative, (I didn't fancy trying to solve derivative = 0 algebraically). First approximation came from looking at the graph.
• Nov 30th 2012, 06:11 AM
kspkido
Re: Finding the minimum value
w8 the d(tanx)/dx is tan^2(x)? lol and the derivative of cot(x)=-cot^2(x)... i cant seem to follow that.....
• Nov 30th 2012, 02:07 PM
gfbrd
Re: Finding the minimum value
Well I know the answer now is 2sqrt(2)-1
Not sure how to get that though =/

derivative of tan is sec^2 and
derivative of cot is csc^2
• Nov 30th 2012, 05:20 PM
x3bnm
Re: Finding the minimum value
Sorry I messed up in my early post. Please disregard my last answer. That's absolutely garbage. Sorry about that.