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avatar+293 

please help need this ASAP

 Nov 24, 2020
 #1
avatar
0

x = 78.46304097º

 

x/2 = 39.23152048º

 

sin x/2 = 0.632455532               cos x/2 = 0.774596669                tan x/2 = 0.81649658

 Nov 24, 2020
 #2
avatar+116126 
+2

To avoid confusion....I'm subbing  theta  for x

 

0° < theta < 90°    means that  x is in the  first quadrant......so theta/2 will also be in the  first  quadrant

 

tan theta = 2sqrt (6)   means that   y  =2sqrt (6) = sqrt (24)   and  x = 1

 

r =  sqrt  ( x^2  + y^2) =   sqrt  ( 1^2  + (sqrt (24))^2 ) =  sqrt (  1 + 24)  = sqrt (25)  = 5

 

cos theta =  x/ r=     1/5

 

sin  (theta/2)  =  sqrt [  ( 1 - cos (theta) )  /2  ] =    sqrt  [ (1 - 1/5) /2 ]  = sqrt  [ (4/5) /2 ] =  sqrt [ 4/10] = (2/10)sqrt (10)  =    sqrt(10)/5

 

cos (theta/2)  =  sqrt  [ (1 + cos (theta)) / 2  ] =    sqrt  [ (1 + 1/5) / 2 ]  = sqrt  [ (6/5)/2 ] = sqrt ( 6 ) /sqrt (10)  =

sqrt (6)sqrt (10) / 10  =  sqrt (60) / 10  =  2sqrt (15) / 10  =  sqrt (15)/5

 

tan (theta/2)  =   1 -cos (theta) / sin (theta)  =   (1 - sqrt (10)/5)  / (sqrt (15)/5)  =  

(5 -sqrt (10)) / sqrt (15)   =   [ 5sqrt (15) -sqrt (10)sqrt (15) ]  /15    =     [ 5sqrt (15)  - sqrt (150 ] / 15

 

 

cool cool cool

 Nov 24, 2020

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