solve sinx/2,cosx/2 & tanx/2 of the following:-
(a)sinx=1/4 & x lies in second quadrant.
draw your triangle
the other side is √15
since we are in QII and sin is positive,
x = -√15
y = 1
r = 4
so, cosx = -√15/4
Now use those values to plug into your half-angle formulas.
Detail
To solve for sin(x/2), cos(x/2), and tan(x/2) when sin(x) = 1/4 and x lies in the second quadrant, follow these steps:
1. Start by finding the value of cos(x) using the Pythagorean Identity:
sin^2(x) + cos^2(x) = 1
We know that sin(x) = 1/4, so substituting the value:
(1/4)^2 + cos^2(x) = 1
1/16 + cos^2(x) = 1
cos^2(x) = 1 - 1/16
cos^2(x) = 16/16 - 1/16
cos^2(x) = 15/16
Taking the square root of both sides:
cos(x) = ±√(15/16)
Since x lies in the second quadrant (where cosine is negative), we take the negative square root:
cos(x) = -√(15/16)
2. Now, let's find sin(x/2):
sin(x/2) = ±√((1 - cos(x)) / 2)
Substituting the value of cos(x) that we already found:
sin(x/2) = ±√((1 - (-√(15/16))) / 2)
sin(x/2) = ±√((1 + √(15/16)) / 2)
Since x lies in the second quadrant (where sine is positive), we take the positive square root:
sin(x/2) = √((1 + √(15/16)) / 2)
3. Finally, let's find tan(x/2):
tan(x/2) = sin(x/2) / cos(x/2)
Substituting the values we found for sin(x/2) and cos(x/2):
tan(x/2) = √((1 + √(15/16)) / 2) / (-√(15/16))
Simplifying the expression, we can rationalize the denominator:
tan(x/2) = -√2 * √(1 + √15) / √15
Now you have the values of sin(x/2), cos(x/2), and tan(x/2) for the given conditions.