sin x = 1 / 4
cos x = ± √ ( 1 - sin ^ 2 x )
In II quadrant cosine is negative so:
cos x = - √ ( 1 - sin ^ 2 x )
cos x = - √ [ 1 - (1 / 4 ) ^ 2 ]
cos x = - √ [ 1 - (1 / 16 ) ]
cos x = - √ [ 16 / 16 - (1 / 16 ) ]
cos x = - √ [ ( 16 - 1 ) / 16 ) ]
cos x = - √ ( 15 / 16 )
cos x = - √15 / 4
Now:
sin ( x / 2 ) = ± √ [ ( 1 - cos x ) / 2 ]
If angle lies in quadrant II half angle lies in quadrant I.
In I quadrant sine is positive so:
sin ( x / 2 ) =√ [ ( 1 - cos x ) / 2 ]
sin ( x / 2 ) = √ [ ( 1 - ( - √15 / 4 ) ) / 2 ]
sin ( x / 2 ) = √ [ ( 1 + √15 / 4 ) / 2 ]
sin ( x / 2 ) = √ [ ( 4 / 4 + √15 / 4 ) / 2 ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / 4 / 2 ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) 4 * 2 ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / ( 2 * 4 ) ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / ( 2 * √ 4 ) ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / ( 2 ) * √ 4 ) ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / ( 2 ) * 2 ) ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / 2 ] / 2
cos ( x / 2 ) = ± √ [ ( 1 + cos x ) / 2 ]
In I quadrant cosine is positive so:
cos ( x / 2 ) =√ [ ( 1 + cos x ) / 2 ]
cos ( x / 2 ) = √ [ ( 1 + ( - √15 / 4 ) ) / 2 ]
cos ( x / 2 ) = √ [ ( 1 - √15 / 4 ) / 2 ]
cos ( x / 2 ) = √ [ ( 4 / 4 - √15 / 4 ) / 2 ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) / 4 / 2 ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) 4 * 2 ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) / ( 2 * 4 ) ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) / ( 2 * √ 4 ) ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) / ( 2 ) * 2 ) ]
cos ( x / 2 ) = √ [ ( 4 - √15 ) / 2 ] / 2
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ √ [ ( 4 + √15 ) / 2 ] / 2 ] / [ √ [ ( 4 - √15 ) / 2 ] / 2 ]
tan ( x / 2 ) = √ ( 4 + √15 ) / √ ( 4 - √15 )
tan ( x / 2 ) = √ [ ( 4 + √15 ) / ( 4 - √15 ) ]
Sin x\2,cos x\2 and tan x\2 for sin x=1\4. x in II quadrant.
3 answers
You also can write this in simplified form:
sin ( x / 2 ) = √ [ ( 4 + √15 ) / 4 * 2 ]
sin ( x / 2 ) = √ ( 4 + √15 ) / √ ( 4 * 2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / ( √4 * √2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / ( 2 * √2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / 2 √2
cos ( x / 2 ) = √ [ ( 4 - √15 ) / 4 * 2 ]
cos ( x / 2 ) = √ ( 4 - √15 ) / √ ( 4 * 2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / ( √4 * √2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / ( 2 * √2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / 2 √2
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ √ ( 4 + √15 ) / 2 √2 ] / [ √ ( 4 - √15 ) / 2 √2 ]
tan ( x / 2 ) = √ ( 4 + √15 ) / √ ( 4 - √15 )
tan ( x / 2 ) = √ [ ( 4 + √15 ) / ( 4 - √15 ) ]
sin ( x / 2 ) = √ [ ( 4 + √15 ) / 4 * 2 ]
sin ( x / 2 ) = √ ( 4 + √15 ) / √ ( 4 * 2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / ( √4 * √2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / ( 2 * √2 )
sin ( x / 2 ) = √ ( 4 + √15 ) / 2 √2
cos ( x / 2 ) = √ [ ( 4 - √15 ) / 4 * 2 ]
cos ( x / 2 ) = √ ( 4 - √15 ) / √ ( 4 * 2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / ( √4 * √2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / ( 2 * √2 )
cos ( x / 2 ) = √ ( 4 - √15 ) / 2 √2
tan ( x / 2 ) = sin ( x / 2 ) / cos ( x / 2 )
tan ( x / 2 ) = [ √ ( 4 + √15 ) / 2 √2 ] / [ √ ( 4 - √15 ) / 2 √2 ]
tan ( x / 2 ) = √ ( 4 + √15 ) / √ ( 4 - √15 )
tan ( x / 2 ) = √ [ ( 4 + √15 ) / ( 4 - √15 ) ]
Belated thanks...
3 answers