# sine inverse

• Oct 10th 2007, 03:25 AM
taurus
sine 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 AM
Jhevon
Quote:

Originally Posted by taurus
Hey
how do i solve this:
sinInverse (sqrt(3/2))

since its range is not between -1 and 1
?

i suppose you meant $\displaystyle \frac {\sqrt{3}}2$

Let $\displaystyle \sin^{-1} \left( \frac {\sqrt{3}}2\right) = x$

$\displaystyle \Rightarrow \sin x = \frac {\sqrt{3}}2$ .........remember the range of arcsine

now, what does x have to be?
• Oct 10th 2007, 03:43 AM
taurus

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 AM
Jhevon
umm...you didn't answer my last question

Quote:

Originally Posted by taurus

secInverse (sqrt(2))
?

i know sec(x)=1/cos(x) but i dont know secInverse and how to do the sqrt(2)

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 AM
taurus
45 correct?
• Oct 10th 2007, 03:51 AM
Jhevon
Quote:

Originally Posted by taurus
45 correct?

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 $\displaystyle \frac {\pi}4$