# Math Help - sine inverse

1. ## sine inverse

Hey
how do i solve this:
sinInverse (sqrt(3/2))

since its range is not between -1 and 1
?

2. 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 $\frac {\sqrt{3}}2$

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

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

now, what does x have to be?

secInverse (sqrt(2))
?

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

4. umm...you didn't answer my last question

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 $\sec^{-1}(\sqrt{2}) = x$

$\Rightarrow \sec x = \sqrt{2}$

$\Rightarrow \frac 1{\cos x} = \sqrt{2}$

$\Rightarrow \cos x = \frac 1{\sqrt{2}}$

Now what does $x$ have to be for this to happen?

I ask you these questions because these are special angles that you SHOULD and MUST know

5. 45 correct?

6. 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 $\frac {\pi}4$
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