Question

I need help; medal to the best answer.
Find the exact value: (problem in the comments)

Answer 1

\[\csc(17\pi/12)\]

Answer 2

\[\csc(x)=\frac{ 1 }{ \sin(x) } \rightarrow \csc \left( \frac{ 17 \pi }{ 12 } \right)=?\]
@InTheTardis

Answer 3

I got to \[(\sqrt{6} +\sqrt{2})/2\] as being what sinx equals. Now what?

Answer 4

well to start with you need to know the quadrant you are working in...

looks like the 3rd....so sin(x) is negative... s csc(x) will also be negative and the 1st quadrant

and

\[\frac{17\pi}{12} = \frac{5\pi}{3} - \frac{\pi}{4}\]

then you can use the difference of 2 angles.

\[\frac{1}{\sin(\frac{5\pi}{3} - \frac{\pi}{4})}\]

Answer 5

\[\frac{ 2 }{ \sqrt{6}-\sqrt{2}}\] I have this. Would I multiply by the conjugate of the denominator to rationalize it? And how did you know to do a difference of squares rather than, say, \[\frac{ 3\pi }{ 4 } + \frac{ 2\pi }{ 3 }\] ?

Answer 6

well that works..

Answer 7

That still doesn't answer my questions.

Answer 8

yes if you want a rational denominator...

you use the difference of 2 squares because the square root of

\[(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2}) = 6 - 2 = 4\]

Answer 9

but the interesting thing is my solution is different

\[\frac{1}{\sin(\frac{3\pi}{4} + \frac{2\pi}{3})}\]

which is

\[\frac{1}{\sin(\frac{3\pi}{4})\cos(\frac{2\pi}{3}) + \cos(\frac{3\pi}{4})\sin(\frac{2\pi}{3})}\]

which becomes

\[\frac{1}{\frac{1}{\sqrt{2}} \times \frac{-1}{2} + \frac{\sqrt{3}}{2} \times - \frac{1}{\sqrt{2}}}\]

which simplifies to

\[- \frac{1}{\frac{1 + \sqrt{3}}{2\sqrt{2}}}\]

dividing by a fraction.. so find the reciprocal an multiply gives

\[-\frac{2\sqrt{2}}{1 + \sqrt{3}}\]

rationalize the denominator

\[- \frac{2 \sqrt{2}}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = -\frac{2\sqrt{2}
- 2\sqrt{6}}{1 - 3} \]

which simplifies to

\[-\frac{(2\sqrt{2} - 2\sqrt{6})}{(-2)} = \sqrt{2} - \sqrt{6} \]

Answer 10

which is a sensible answer as its negative... 3rd quadrant where its negative

and larger then 1 which occurs with csc

Answer 11

Thank you. I think I have the hang of it now.