Suppose Start Fraction pi Over 2 End Fraction less-than theta less-than pi and sine ⁡ theta equals Start Fraction 2 Over 5 End Fraction period Evaluate cosine ⁡ theta.

I am going to guess that you meant:

sin Ø = 2/5, and π/2 < Ø ≤ π
or , even more simply stated ....

sin Ø = 2/5 and Ø is in the second quadrant

make a sketch of your right-angled triangle, with
y = 2, and r = 5
By Pythagoras: x^2 + y^2 = 5^2
x^2 + 4 = 25
x = -√21 in the 2nd quad

so cosØ = -√21/5

To evaluate cosine(theta), we need to find the exact value of theta within the given interval and use the relationship between sine and cosine functions.

Given that Start Fraction pi/2 < theta < pi and sine(theta) = 2/5, we can use the Pythagorean identity to find the value of cosine(theta).

The Pythagorean identity states that for any angle theta in a right triangle,

sin^2(theta) + cos^2(theta) = 1.

Since we know that sine(theta) equals 2/5, we can substitute this value into the equation:

(2/5)^2 + cos^2(theta) = 1.

Simplifying the equation:

4/25 + cos^2(theta) = 1.

Next, we can solve for cos(theta) by isolating it:

cos^2(theta) = 1 - 4/25,

cos^2(theta) = 21/25.

To find cos(theta), take the square root of both sides:

cos(theta) = ±sqrt(21/25).

Since theta lies between pi/2 and pi, which is the second quadrant, the cosine function is negative. Therefore, we take the negative square root:

cos(theta) = -sqrt(21/25).

Hence, the value of cosine(theta) is -sqrt(21/25).