Question

Sin x\2,cos x\2 and tan x\2 for sin x=1\4. x in II quadrant.

Answer 1

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 ) ]

Answer 2

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 ) ]

Answer 3

Belated thanks...