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C3 Trig

Given that x= cosec(Y) + cot (Y) find value of Y that satisfy the equation x+1/x=4 giving your solutions in the interval 0<Y<360
Can you show what you've tried, and show where you got stuck?
Reply 2
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Niamh99
OP
4
I tried to sub in cosec(Y) + Cot(Y) into x+1/x=4 but I wasn’t sure if this was the right method. Do you think this is the right method? Or is there something else I should do first?
Original post by Niamh99
I tried to sub in cosec(Y) + Cot(Y) into x+1/x=4 but I wasn’t sure if this was the right method. Do you think this is the right method? Or is there something else I should do first?


I feel like you should arrange x+1x=4\dfrac{x+1}{x} =4 into another x= expression so you can do x=x, i also feel like you should rearrange cosec(y)+cot(y)cosec(y)+cot(y) such that it's only 1 trig function. that makes things easier right?
Reply 4
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Niamh99
OP
4
Original post by will'o'wisp2
I feel like you should arrange \dfrac{x+1}{x} =4. into another x= expression so you can do x=x, i also feel like you should rearrange cosec(y)+cot(y) such that it's only 1 trig function. that makes things easier right?
So if I rearranged cosec(Y) + cot(Y) = x to [1+cos(Y)/sin(Y)] do you think i could put that into x+ (1/x) = 4 ?
Original post by Niamh99
I tried to sub in cosec(Y) + Cot(Y) into x+1/x=4 but I wasn’t sure if this was the right method. Do you think this is the right method? Or is there something else I should do first?


You can simplify csc⁡(y)+cot⁡(y)+1csc⁡(y)+cot⁡(y)\csc(y) + \cot(y) + \frac 1 {\csc(y)+\cot(y)} quite nicely. Try writing it as a single fraction. Hint once you've got there, cot⁡2(y)+1=?\cot^2(y) + 1 = ? (use sin⁡2(y)+cos⁡2(y)≡1\sin^2(y)+\cos^2(y) \equiv 1). You should then be able to cancel a factor.
(edited 6 years ago)
Original post by Niamh99
Original post by will'o'wisp2
I feel like you should arrange \dfrac{x+1}{x} =4. into another x= expression so you can do x=x, i also feel like you should rearrange cosec(y)+cot(y) such that it's only 1 trig function. that makes things easier right?
So if I rearranged cosec(Y) + cot(Y) = x to [1+cos(Y)/sin(Y)] do you think i could put that into x+ (1/x) = 4 ?


you missed the
lol nvm

but where does that come from tho?

cosec=1/sin and cot is 1/tan or cos/sin

ah i see ambiguous notiation lol so it's x+1x=4x + \dfrac{1}{x} =4

in which case then i think you should complete the square on x+1x=4x + \dfrac{1}{x} =4 and rearrange for x= and you should probably still rearrange the trig stuff into a single trig function
[QUOTE="Niamh99;74039286"]
Original post by will'o'wisp2
I feel like you should arrange \dfrac{x+1}{x} =4. into another x= expression so you can do x=x, i also feel like you should rearrange cosec(y)+cot(y) such that it's only 1 trig function. that makes things easier right?
So if I rearranged cosec(Y) + cot(Y) = x to [1+cos(Y)/sin(Y)] do you think i could put that into x+ (1/x) = 4 ?


Original post by _gcx
You can simplify csc⁡(y)+cot⁡(y)+1csc⁡(y)+cot⁡(y)\csc(y) + \cot(y) + \frac 1 {\csc(y)+\cot(y)} quite nicely. Try writing it as a single fraction. Hint once you've got there, cot⁡2(y)+1=?\cot^2(y) + 1 = ? (use sin⁡2(y)+cos⁡2(y)≡1\sin^2(y)+\cos^2(y) \equiv 1). You should then be able to cancel a factor.


^^ this guy , use this
Reply 8
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Niamh99
OP
4
Thank you guys, I’ve solved it now :smile:

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