Hey

how do i solve this:

sinInverse (sqrt(3/2))

since its range is not between -1 and 1

?

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- Oct 10th 2007, 03:25 AMtaurussine inverse
Hey

how do i solve this:

sinInverse (sqrt(3/2))

since its range is not between -1 and 1

? - Oct 10th 2007, 03:34 AMJhevon
- Oct 10th 2007, 03:43 AMtaurus
oops yea, it was this other question i had problems with:

secInverse (sqrt(2))

?

i know sec(x)=1/cos(x) but i dont know secInverse and how to do the sqrt(2) - Oct 10th 2007, 03:47 AMJhevon
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 - Oct 10th 2007, 03:49 AMtaurus
45 correct?

- Oct 10th 2007, 03:51 AMJhevon
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 $\displaystyle \frac {\pi}4$