Question

reference angles... help me finish this please? (Problem in comments)

Answer 1

\[\csc (-60)\]

Answer 2

use SOH CAH TOA :/
\(sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)

to get the cosine
I'd think that'd be obvious

Answer 3

I did do that... I don't know how to continue though.

Answer 4

well.. what did you get by using the cosine identity then?

Answer 5

1/-sqrt-3/2

Answer 6

ohh hmmm rats... is cosecant... kinda misread there, thought it was cosine

Answer 7

hmmm ok... so.. .your cosecant is correct.... but we do know from the getgo is a -60 degree angle, so... what are you asked to find again?

Answer 8

csc -60

Answer 9

well, you found it :)

Answer 10

how would I simplify it though?

Answer 11

It's better to use



Now we have to simplify

\(\Large\bf -\frac{2}{\sqrt{3}}\)

The easiest way to do that is to multiply the numerator and denominator by sqrt(3) so that way we can "rationalize" the denominator.

\(\Large\bf -\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=?\)

Answer 12

I'm not allowed to do that :(