Please find sin (alpha/2) and cos (alpha/2)

sec alpha = 3, -pi/2 < alpha < 0

To find sin(alpha/2) and cos(alpha/2), we need to use the given information that sec(alpha) = 3.

Recall that sec(alpha) is the reciprocal of the cosine function. So, we can find the cosine value by taking the reciprocal of the given sec(alpha) value:

cos(alpha) = 1 / sec(alpha)
= 1 / 3

Now, to find sin(alpha), we can use the Pythagorean identity which states that sin^2(alpha) + cos^2(alpha) = 1. We already know cos(alpha) from the previous step, so we can rearrange the equation to find sin(alpha):

sin^2(alpha) = 1 - cos^2(alpha)
= 1 - (1/3)^2
= 1 - 1/9
= 8/9

Taking the square root of both sides:

sin(alpha) = sqrt(8/9)
= sqrt(8) / sqrt(9)
= (sqrt(8) / 3)

Finally, to find sin(alpha/2) and cos(alpha/2), we can use the half-angle formula:

sin(alpha/2) = sqrt((1 - cos(alpha)) / 2)
= sqrt((1 - 1/3) / 2)
= sqrt(2/6)
= sqrt(1/3)
= 1/sqrt(3)
= sqrt(3) / 3

cos(alpha/2) = sqrt((1 + cos(alpha)) / 2)
= sqrt((1 + 1/3) / 2)
= sqrt(4/6)
= sqrt(2/3)
= sqrt(2) / sqrt(3)
= sqrt(6) / 3

Therefore, sin(alpha/2) = sqrt(3)/3 and cos(alpha/2) = sqrt(6)/3.