alright...so here I am trying to figure out a step in my Calculus problem, and I somehow need to go from sec(squared)[pi/4] = (Square root 2)(squared) = 2.....could anyone explain this?

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- May 15th 2008, 01:19 PMCALsunshineHelp with Calc problem....
alright...so here I am trying to figure out a step in my Calculus problem, and I somehow need to go from sec(squared)[pi/4] = (Square root 2)(squared) = 2.....could anyone explain this?

- May 15th 2008, 01:23 PMMoo
Hello,

$\displaystyle \begin{aligned} \sec^2 \frac{\pi}{4} &=\frac{1}{\cos^2 \frac{\pi}{4}} \\

&=\frac{1}{\left(\frac{\sqrt{2}}{2}\right)^2} \end{aligned}$

$\displaystyle \begin{aligned} \sec^2 \frac{\pi}{4} &=\left(\frac{2}{\sqrt{2}}\right)^2 \\

&=\left(\frac{\sqrt{2}\cdot {\color{red}\sqrt{2}}}{\color{red} \sqrt{2}}\right)^2 \\

&=(\sqrt{2})^2 \end{aligned}$ - May 15th 2008, 01:24 PMtopsquark
- May 15th 2008, 01:24 PMChris L T521
$\displaystyle \sec^{2}\left(\frac{\pi}{4}\right)$

Recall that $\displaystyle \sec u=\frac{1}{\cos u}$ and $\displaystyle \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$. Thus, we see then that $\displaystyle sec\left(\frac{\pi}{4}\right)=\frac{1}{\frac{1}{\s qrt{2}}}=\sqrt{2}$.

Therefore, $\displaystyle \sec^{2}\left(\frac{\pi}{4}\right)=\left(\sqrt{2}\ right)^2=2$.

Hope that answers your question!! :D - May 15th 2008, 01:36 PMCALsunshine

Thanks guys!!!! I totally forgot to use my unit circle....that would have made my life a lot easier! :)