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a simple trigonomertic equation gone totally insane/////

sin²x + cot²x = 4 [0° ≤ x ≤ 360̇̇°]

Your simple effort is Kindly appreciated!!! :biggrin:
Write the cotx squared in terms of sin only and simplify.
Original post by UdipR1
sin²x + cot²x = 4 [0° ≤ x ≤ 360̇̇°]

Your simple effort is Kindly appreciated!!! :biggrin:


Here it is

sin^2(x)/8-(3 cos^2(x))/4+2 cot^2(x)-(9 csc^2(x))/8+1/8 cos^2(x) cot^2(x)-2

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UdipR1
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Original post by physicst
Here it is

sin^2(x)/8-(3 cos^2(x))/4+2 cot^2(x)-(9 csc^2(x))/8+1/8 cos^2(x) cot^2(x)-2



I am just in my high school now,,So I am not getting what you did .I used simplication and trigonometric 's formulas but it got me nowhere.Anyways,thanks for your help. :smile:
(edited 9 years ago)
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UdipR1
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Original post by Phoebe Buffay
Write the cotx squared in terms of sin only and simplify.


I tried but It didn't do any good,thanx for your suggestion by the way.
Original post by UdipR1
I am just in my high school now,,So I am not getting what you did .I used simplication and trigonometric 's formulas but it got me nowhere.Anyways,thanks for your help. :smile:



This is just the expansion of your expression in more simplified version. In figure value of your function is plotted in between 0<x<2Pi. What actually you want to do . you want to simplify it? or just want to see its behaviour in range 0<x<2Pi. The figure contain its behaviour. 360 is divided into six parts and mention along x-axis and value of function on y-axis.

Have a nice day

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