Hey
how do i solve this:
sinInverse (sqrt(3/2))
since its range is not between -1 and 1
?
umm...you didn't answer my last question
Let $\displaystyle \sec^{-1}(\sqrt{2}) = x$
$\displaystyle \Rightarrow \sec x = \sqrt{2}$
$\displaystyle \Rightarrow \frac 1{\cos x} = \sqrt{2}$
$\displaystyle \Rightarrow \cos x = \frac 1{\sqrt{2}}$
Now what does $\displaystyle x$ have to be for this to happen?
I ask you these questions because these are special angles that you SHOULD and MUST know
yes (this is for the cosine question, correct? i asked you two questions, you should be specific as to which you are answering). that is in degrees. you should tell these answers in radians i believe, unless otherwise instructed. that's what they use in college math. so the answer would be $\displaystyle \frac {\pi}4$