If x=6csc(theta) in the equation sqrt(x^2-36) how can you find sin and cos of this? All assuming 0

Question

If x=6csc(theta) in the equation sqrt(x^2-36) how can you find sin and cos of this? All assuming 0<theta<pi/2

jdoe0001 · Answer

is there any context to it?  $\bf \sqrt{6csc^2(\theta)-36}$   doesn't have to have a sine or cosine per se

anonymous · Answer

Not really, it says "Use the given substitution to express the given radical expression as a trig function without radicals" but it's asking for$$\sin(\theta) and \cos(\theta). $$ I just dont know how to find out what theta is from what it has given.

jdoe0001 · Answer

so it needs to be expressed in terms of sine AND/OR cosine?

anonymous · Answer

The question asks for both sin and cos of theta.

jdoe0001 · Answer

hmmm

jdoe0001 · Answer

not sure I follow....    can you post a quick screenshot of the material?

anonymous · Answer

attachment

jdoe0001 · Answer

I can see AND/OR happening.... and once you get either, you can expand to the other

anonymous · Answer

Sorry for the confusion, I just don't really understand the question? I don't know how to simplify from the original equation to find theta and take that to finding sin/cos

jdoe0001 · Answer

well.... I can see what they're asking... the sine and cosine of the angle.  which is in the 1st Quadrant

however I doin't see any " = " equals signs there to equate it against

we could simplify it to be cosine based or sine based.... but not equate it to any angle

jdoe0001 · Answer

so.. there isn't really any EQUATion

anonymous · Answer

So how can you simplify the original square root if there isnt any equation? That's what is confusing me so much, I don't know how to solve it when there's nothing to set it equal to...

jdoe0001 · Answer

exactly, there's nothing equals to
we can surely simplify it some.... but that won't equate anything, or any angle for that matter

jdoe0001 · Answer

I assume there might have been a =0 somewhere, but since to have been left out, dunno

anonymous · Answer

What is theta when it equal zero?

jdoe0001 · Answer

well, if use use $\bf 0=\sqrt{[6csc(\theta)]^2-36}$   we'd be assuming that's so, but we really dunno

jdoe0001 · Answer

if we use I meant anyhow

so... is a stretch there..... some  = was left out it seems

anonymous · Answer

apparently this is what they wanted? Not sure how that works out. Thanks for your help!

jdoe0001 · Answer

looks incomplete.... or maybe there's an example somewhere in your material on what they were doing