Asked by Amber
                Find sin x/2, cos x/2, and tan x/2 from the given information. 
[1]. sin x = 5/13, 0° < x < 90°
1). sin x/2 =________.
2). cos x/2 =________.
3). tan x/2 =________.
[Note: Plz help with these 3]
            
            
        [1]. sin x = 5/13, 0° < x < 90°
1). sin x/2 =________.
2). cos x/2 =________.
3). tan x/2 =________.
[Note: Plz help with these 3]
Answers
                    Answered by
            Bosnian
            
    Use Trigonometric Identities :
sin ( x / 2 ) = + OR - sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = + OR - sqroot [ ( 1 + cos x ) / 2 ]
0° < x < 90° meen x lies in Quadrant I
In Quadrant I sin x and cos x are positive so :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
If sin x = 5 / 13 then :
cos x = sqroot [ 1 - ( sin x ) ^ 2 ]
cos x = sqroot [ 1 - ( 5 / 13 ) ^ 2 ] =
sqroot [ 1 - 25 / 169 ] =
sqroot [ 169 / 169 - 25 / 169 ] =
sqroot [ 144 / 169 ] = 12 / 13
cos x = 12 / 13
Now :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
sin ( x / 2 ) = sqroot [ ( 1 - 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 - 12 / 13 ) / 2 ] =
sqroot [ ( 1 / 13 ) / 2 ] =
sqroot [ 1 / ( 2 * 13 ) ] =
sqroot ( 1 / 26 ) = 1 / sqroot ( 26 )
sin ( x / 2 ) = 1 / sqroot ( 26 )
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 + 12 / 13 ) / 2 ] =
sqroot [ ( 25 / 13 ) / 2 ] =
sqroot [ 25 / ( 2 * 13 ) ] =
sqroot ( 25 / 26 ) = 5 / sqroot ( 26 )
cos ( x / 2 ) = 5 / sqroot ( 26 )
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ 1 / sqroot ( 26 ) ] / [ 5 / sqroot ( 26 ) ] =
[ 1 * sqroot ( 26 ) ] / [ 5 * sqroot ( 26 ) ] = 1 / 5
tan ( x / 2 ) = 1 / 5
    
sin ( x / 2 ) = + OR - sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = + OR - sqroot [ ( 1 + cos x ) / 2 ]
0° < x < 90° meen x lies in Quadrant I
In Quadrant I sin x and cos x are positive so :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
If sin x = 5 / 13 then :
cos x = sqroot [ 1 - ( sin x ) ^ 2 ]
cos x = sqroot [ 1 - ( 5 / 13 ) ^ 2 ] =
sqroot [ 1 - 25 / 169 ] =
sqroot [ 169 / 169 - 25 / 169 ] =
sqroot [ 144 / 169 ] = 12 / 13
cos x = 12 / 13
Now :
sin ( x / 2 ) = sqroot [ ( 1 - cos x ) / 2 ]
sin ( x / 2 ) = sqroot [ ( 1 - 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 - 12 / 13 ) / 2 ] =
sqroot [ ( 1 / 13 ) / 2 ] =
sqroot [ 1 / ( 2 * 13 ) ] =
sqroot ( 1 / 26 ) = 1 / sqroot ( 26 )
sin ( x / 2 ) = 1 / sqroot ( 26 )
cos ( x / 2 ) = sqroot [ ( 1 + cos x ) / 2 ]
cos ( x / 2 ) = sqroot [ ( 1 + 12 / 13 ) / 2 ] =
sqroot [ ( 13 / 13 + 12 / 13 ) / 2 ] =
sqroot [ ( 25 / 13 ) / 2 ] =
sqroot [ 25 / ( 2 * 13 ) ] =
sqroot ( 25 / 26 ) = 5 / sqroot ( 26 )
cos ( x / 2 ) = 5 / sqroot ( 26 )
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ 1 / sqroot ( 26 ) ] / [ 5 / sqroot ( 26 ) ] =
[ 1 * sqroot ( 26 ) ] / [ 5 * sqroot ( 26 ) ] = 1 / 5
tan ( x / 2 ) = 1 / 5
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