Question

\[sec\left(cos^{-1}\frac 12\right)\]
\[cos\alpha=\frac 12\]
\[\alpha=\frac {\pi} 3\]
where from here?

Answer 1

oh is it just
\[sec(\frac \pi 3)\]

Answer 2

\[\large \sec\left(\arccos\frac{1}{2}\right) \qquad = \qquad \sec\left(\frac{\pi}{3}\right) \qquad = \qquad ?\]

Yah I think you've got the right idea :)

Answer 3

2

Answer 4

Mmmmm yah sounds good c:

Answer 5

yay....let me try another one

Answer 6

Make sure you're comfortable with this notation \(\large \cos^{-1}x\). The -1 is never a power. It always represents the inverse function when we're dealing with trig.

Answer 7

\[\large \cos^{-1}x \neq \frac{1}{\cos x}\]

Answer 8

Yep makes sense :)