# Help with Calc problem....

• May 15th 2008, 01:19 PM
CALsunshine
Help 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 PM
Moo
Hello,

Quote:

Originally Posted by CALsunshine
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?

\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}

\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 PM
topsquark
Quote:

Originally Posted by CALsunshine
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?

$sec^2 \left ( \frac{\pi}{4} \right )$

$= \frac{1}{cos^2 \left ( \frac{\pi}{4} \right ) }$

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

$= \frac{1}{\frac{1}{2}}$

$= 2$

-Dan
• May 15th 2008, 01:24 PM
Chris L T521
Quote:

Originally Posted by CALsunshine
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?

$\sec^{2}\left(\frac{\pi}{4}\right)$

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

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