Question

jl

Answer 1

First we need to get a picture in our heads: I arbitrarily chose that the radius of the circle is one.

Answer 2

\[\csc x=\frac{ 1 }{ \cos x }=9 \rightarrow \cos x=\frac{ 1 }{ 9 }\]So:Here I think you need to use some Half angle dormulas.

Answer 3

First half-angle formula:

\[\cos x=2 \cos ^{2} \frac{ x }{ 2 }-1 \rightarrow \cos \frac{ x }{ 2 }=\sqrt{\frac{ \cos x +1 }{ 2 }}=\sqrt{\frac{ \frac{ 1 }{ 9 }+1 }{ 2 }}\]Now you do the same with this formulas: \[cosx=1-2 \sin ^{2} \frac{ x }{ 2 }\]For the last one, just divide the sin and the cos

Answer 4

Well, I had that doubt because you are right. The problem is that the initial condition is cscx=9, not -9

Answer 5

Well, acording to my equation, cos should be sqrt 10/18

Answer 6

Oh yeah you are right

Answer 7

You just need to apply the half.angle formula's I gave for what you got for cosx

Answer 8

jhalt post your work for cos x/2

(I assume you already have sin x/2 ?)

Answer 9

yes, I understand... but what did you do ?

Answer 10

isn't cos x = - sqrt(80)/9
?

Answer 11

yes
you are doing 1 + cos(x)
as the first step
so 1 + - sqrt(80)/9

Answer 12

looks good

Answer 13

tan is sin / cos

Answer 14

yes, that all is correct