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

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Thread starter 5 years ago
#1
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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5 years ago
#2
(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?
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5 years ago
#3
(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
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5 years ago
#4
I think

Cosec2x = 1/sin2x= 1/2sinxcosx

Then sub in p/q for sinx
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5 years ago
#5
(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 in terms of and how would you find in terms of and

Hint

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5 years ago
#6
(Original post by Khallil)
If the question is asking you to express in terms of and how would you find in terms of and

Hint

Is it 1-p/q?
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5 years ago
#7
(Original post by Vorsah)
Is it 1-p/q?
Not quite.

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5 years ago
#8
(Original post by Khallil)
Not quite.

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?
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5 years ago
#9
(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?
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5 years ago
#10
(Original post by Khallil)
Okay,

But I dunno what to do from cos^2x= 1- p^2/q^2?
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5 years ago
#11
(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).
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5 years ago
#12
(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
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5 years ago
#13
(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
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5 years ago
#14
(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 ?
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5 years ago
#15
(Original post by Vorsah)
Cosx= (q-p)/q ?
It's almost as if you didn't read my post. Try again.
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5 years ago
#16
(Original post by Vorsah)
Cosx= (q-p)/q ?
lol
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5 years ago
#17
(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?
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5 years ago
#18
(Original post by keromedic)
lol
I wouldn't lol if you got a question wrong.
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5 years ago
#19
(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.
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Thread starter 5 years ago
#20
Thank you so much for your help. I think I've got it now. 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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