Question

how do I evaluate arccsc root2?
Please help me

Answer 1

\[arccsc(\sqrt{2})=y \\ \sqrt{2}=\csc(y) \\ \sqrt{2}=\frac{1}{\sin(y)} \\ \frac{1}{\sqrt{2}}=\sin(y) \\ \arcsin(\frac{1}{\sqrt{2}}) =y \\ \text{ can you say what } \arcsin(\frac{1}{\sqrt{2}}) \text{ equals ? }\]

Answer 2

notice another way to write
\[ \frac{1}{\sqrt{2}} = \frac{\sqrt{2} }{2} \]
you want to find the angle x , when
\[ \sin x = \frac{\sqrt{2} }{2} \]

this is one of the angles you should memorize

Answer 3

The thing is, I'm supposed to figure out how to put that in to something like pi/3, how would I do that?

Answer 4

hmm look at your Unit Circle :)
or use a calculator for that matter, that'd work as well

Answer 5

Do I need a graphing calculator for this?

Answer 6

hmmm hold the mayo
did you mean btw \(\Large cos^{-1}(\sqrt{2})\) ?

Answer 7

No the problem just said to evaluate arccsc ( 1/ \sqrt2)

Answer 8

graphing calculator is not necessary

Answer 9

hmm 1/? where did that come from

Answer 10

u may look at frekel's method at the top , or just study the unit circle as stated

Answer 11

well... still , very simple as freckles showed above
that angle unit, would be in your Unit Circle

Answer 12

There isn't a circle stated with the question, what I said is all that there was

Answer 13

unit circle is the basic topic to know in trignometry.

https://www.mathsisfun.com/geometry/unit-circle.html

Answer 14

so something like this?

Answer 15

it looks like a pizza

Answer 16

Well you get what I'm going for right?

Answer 17

it s not upto 9pi its upto 0 to 360\((2\pi)\)

Answer 18

in unit circle u need to learn the basic values of angles of \(0,30,45,90\) degrees of \(\sin,\tan \cos\) etc.

Answer 19

well...\(\bf cos^{-1}\left( \cfrac{1}{\sqrt{2}} \right)\qquad thus
\\ \quad \\

\cfrac{1}{\sqrt{2}} \cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{1\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}\implies \cfrac{\sqrt{2}}{\sqrt{2^2}}\implies \cfrac{\sqrt{2}}{2}\qquad thus
\\ \quad \\
cos^{-1}\left( \cfrac{1}{\sqrt{2}} \right)\iff cos^{-1}\left( \cfrac{\sqrt{2}}{2} \right)\impliedby
\begin{array}{llll}
\textit{the angle whose cosine}\\
is\quad \cfrac{\sqrt{2}}{2}
\end{array}\)