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sinx=p/q, find cosec2x in terms of p and q

Please could someone help me with this question. I just can't figure it out. Given that sinx=p/q, 0<x<pi/2 and p>0 and q>0, find cosec2x in terms of p and q.

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Reply 1
Original post by madnutcase434
Please could someone help me with this question. I just can't figure it out. Given that sinx=p/q, 0<x<pi/2 and p>0 and q>0, find cosec2x in terms of p and q.


What have you done so far?
Original post by madnutcase434
Please could someone help me with this question. I just can't figure it out. Given that sinx=p/q, 0<x<pi/2 and p>0 and q>0, find cosec2x in terms of p and q.


cosec2x=1/sin2x so just find sinx and use the double angle identity to get sin 2x
Reply 3
I think

Cosec2x = 1/sin2x= 1/2sinxcosx

Then sub in p/q for sinx
Original post by Vorsah
I think

Cosec2x = 1/sin2x= 1/2sinxcosx

Then sub in p/q for sinx


If the question is asking you to express csc2x\csc 2x in terms of pp and qq how would you find cosx\cos x in terms of pp and q ?q \ ?

Hint

Reply 5
Original post by Khallil
If the question is asking you to express csc2x\csc 2x in terms of pp and qq how would you find cosx\cos x in terms of pp and q ?q \ ?

Hint



Is it 1-p/q?
Original post by Vorsah
Is it 1-p/q?


Not quite.

sinx=pq    sin2x=p2q2\sin x = \dfrac{p}{q} \implies \sin^2 x = \dfrac{p^2}{q^2}
Reply 7
Original post by Khallil
Not quite.

sinx=pq    sin2x=p2q2\sin x = \dfrac{p}{q} \implies \sin^2 x = \dfrac{p^2}{q^2}


I thought:

Sinx= p/q >>> sin^2x= (p/q)^2

So

Cos^2x = 1 - (p/q)^2

Cosx = 1-p/q

I don't know where I went wrong?
Original post by Vorsah
I thought:

Sinx= p/q >>> sin^2x= (p/q)^2

So

Cos^2x = 1 - (p/q)^2

Cosx = 1-p/q

I don't know where I went wrong?


cos2x=1p2q2⇏cosx=1pq\cos^2 x = 1 - \dfrac{p^2}{q^2} \not \Rightarrow \cos x = 1 - \dfrac{p}{q}
Reply 9
Original post by Khallil
cos2x=1p2q2⇏cosx=1pq\cos^2 x = 1 - \dfrac{p^2}{q^2} \not \Rightarrow \cos x = 1 - \dfrac{p}{q}


Okay,

But I dunno what to do from cos^2x= 1- p^2/q^2?
Original post by Vorsah
Okay,

But I dunno what to do from cos^2x= 1- p^2/q^2?


Find a common denominator for the RHS and take the square root of the equality. Then substitute what you have for sin(x) and cos(x) into the original expression for cosec(2x).
Reply 11
Original post by Khallil
Find a common denominator for the RHS and take the square root of the equality. Then substitute what you have for sin(x) and cos(x) into the original expression for cosec(2x).


don't forget to put "plus or minus" when you do a square root operation
Reply 12
Original post by the bear
don't forget to put "plus or minus" when you do a square root operation


The question gives a restricted range for x, which should help to fix which sign needs to be taken :smile:
Reply 13
Original post by Khallil
Find a common denominator for the RHS and take the square root of the equality. Then substitute what you have for sin(x) and cos(x) into the original expression for cosec(2x).


Cosx= (q-p)/q ?
Original post by Vorsah
Cosx= (q-p)/q ?


It's almost as if you didn't read my post. Try again.
Original post by Vorsah
Cosx= (q-p)/q ?

lol
Reply 16
Original post by Khallil
It's almost as if you didn't read my post. Try again.


Is it cosx= sqrt (q^2-p^2)/q^2?
Reply 17
Original post by keromedic
lol


I wouldn't lol if you got a question wrong.
(edited 10 years ago)
Original post by Vorsah
I wouldn't lol if you got a question wrong.

I wasn't loling at that.
I was loling because you appeared to have ignored Khalil's advice.
Reply 19
Thank you so much for your help. I think I've got it now.:biggrin: so if cos2x=1-(p2/q2), then cosx=((q2-p2)1/2)/q2 (with some rearranging) and then substitute that along with p/q (=sinx) into 1/(2sinxcosx) and voila!

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