# Differential trig problem

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• May 1st 2009, 02:50 PM
slaypullingcat
Differential trig problem
Hi, I'm obviously missing something here as I am only getting further awway from the answer.

show that d/dx (cos x + sin x/ cos x - sin x) = sec^2(pi/4 +x)

Thanks in advance
• May 1st 2009, 03:33 PM
skeeter
Quote:

Originally Posted by slaypullingcat
Hi, I'm obviously missing something here as I am only getting further awway from the answer.

show that d/dx (cos x + sin x/ cos x - sin x) = sec^2(pi/4 +x)

Thanks in advance

let's do something with the secant mess first ...

$\displaystyle \sec^2\left(\frac{\pi}{4}+x\right) =$

$\displaystyle 1 + \tan^2\left(\frac{\pi}{4}+x\right) =$

$\displaystyle 1 + \left[ \frac{\tan\left(\frac{\pi}{4}\right) + \tan{x}}{1 - \tan\left(\frac{\pi}{4}\right)\tan{x}}\right]^2 =$

$\displaystyle 1 + \left[\frac{1 + \tan{x}}{1 - \tan{x}}\right]^2 =$

$\displaystyle 1 + \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =$

$\displaystyle \left[\frac{\cos{x}-\sin{x}}{\cos{x}-\sin{x}}\right]^2+ \left[\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}}\right]^2 =$

$\displaystyle \frac{(\cos{x}-\sin{x})^2 + (\cos{x}+\sin{x})^2}{(\cos{x}-\sin{x})^2}$

now take the derivative of the left side and see if it matches up.
• May 1st 2009, 03:38 PM
slaypullingcat
thank you very much, that was really easy to follow.