Sin x\2,cos x\2 and tan x\2 for sin x=1\4. x in II quadrant.
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 ) ]
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 ) ]
Belated thanks...
To find the values of sin(x/2), cos(x/2), and tan(x/2) when sin(x) = 1/4 and x is in the second (II) quadrant, we can follow the steps below:
Step 1: Find the value of cos(x) using the given information.
Since sin(x) = 1/4 and x is in the second quadrant, we know that sin(x) is positive in the second quadrant. So sin(x) = 1/4 means that the opposite side of the right triangle has a length of 1, and the hypotenuse has a length of 4. Using the Pythagorean theorem, we can find the length of the adjacent side (cos(x)):
cos(x) = √(1 - sin^2(x)) = √(1 - (1/4)^2) = √(1 - 1/16) = √(15/16) = √15/4 = √15/2.
Step 2: Use the half-angle formulas to find the values of sin(x/2), cos(x/2), and tan(x/2) using the value of cos(x) obtained in Step 1.
The half-angle formulas are:
sin(x/2) = ±√[(1 - cos(x)) / 2]
cos(x/2) = ±√[(1 + cos(x)) / 2]
tan(x/2) = sin(x/2) / cos(x/2)
Since x is in the second quadrant, we know that sin(x/2) and cos(x/2) will both be positive.
Substituting the value of cos(x) from Step 1 into the half-angle formulas:
sin(x/2) = √[(1 - √15/2) / 2]
cos(x/2) = √[(1 + √15/2) / 2]
tan(x/2) = sin(x/2) / cos(x/2)
Now you can simplify the expressions further if needed.
Note: The values of sin(x/2), cos(x/2), and tan(x/2) will be approximations since √15 is an irrational number.