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2sec(x+60)=5sec(x-20)

 Aug 11, 2016
 #1
avatar+130466 
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2sec(x+60)=5sec(x-20)

 

2 /cos (x + 60)  = 5 / cos ( x - 20)

 

2 / [ cosxcos60 - sinxsin60]  =  5 / [ cosxcos20 + sinxsin20]

 

2  [ cosxcos20 + sinxsin20]  =  5 [ cosxcos60 - sinxsin60]

 

2 [ cosx cos20 + sinxsin20]  = 5 [ (cosx)/2 - (sqrt(3)*sinx) /2 ]

 

4[cosxcos20 + sinxsin20]  = 5 [cosx - sqrt(3)*sinx]

 

sinx [ 4sin20 + 5sqrt(3)]  = cosx [ 5 - 4cos20]

 

sinx / cosx   =   [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)]

 

tanx  = [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)]

 

arctan (  [ 5 - 4cos20] /  [ 4sin20 + 5sqrt(3)])  = x =  about 7.056° + n*360°     where n is an integer

 

Here's a graph :  https://www.desmos.com/calculator/oqxckkdfx2

 

 

 

cool cool cool

 Aug 11, 2016
 #2
avatar+26396 
0

2sec(x+60)=5sec(x-20)

 

2sec(x+60)=5sec(x20)We substitute x=y202sec(y20+60)=5sec(y2020)2sec(y+40)=5sec(y40)|sec(ϕ)=1cos(ϕ)2cos(y+40)=5cos(y40)cos(y40)cos(y+40)=52|cos(y40)=cos(y)cos(40)+sin(y)sin(40)cos(y)cos(40)+sin(y)sin(40)cos(y+40)=52|cos(y+40)=cos(y)cos(40)sin(y)sin(40)cos(y)cos(40)+sin(y)sin(40)cos(y)cos(40)cos(y)cos(40)sin(y)sin(40)cos(y)cos(40)=521+tan(y)tan(40)1tan(y)tan(40)=522[ 1+tan(y)tan(40) ]=5[ 1tan(y)tan(40) ]2+2tan(y)tan(40)=55tan(y)tan(40)7tan(y)tan(40)=3tan(y)=371tan(40)|y=x+20tan(x+20)=371tan(40)x+20=arctan(371tan(40))±n180x=20+arctan(371tan(40))±n180x=20+arctan(0.51075153968)±n180x=20+27.0557431286±n180x=7.0557431286±n180nN

 

laugh

 Aug 12, 2016
edited by heureka  Aug 12, 2016

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