If cosx=sqrt{3}/2 and −pi/2<=x<=0 then the exact value of x, in radians
The range for x, -pi/2 <= x <=0, restricts x to the 4th quadrant.
What angle x in the 4th quadrant has the property that cos(x) = sqrt(3)/2?
To find the exact value of x, we need to determine the angle that satisfies the given condition cos(x) = √3/2 in the given interval -π/2 ≤ x ≤ 0.
We know that the cosine function represents the ratio of the adjacent side to the hypotenuse in a right triangle. Since cos(x) = √3/2, we can draw a right triangle where the adjacent side is √3 and the hypotenuse is 2.
Now let's look at the unit circle. In the first quadrant (0 ≤ x ≤ π/2), the cosine value is positive, but in the second quadrant (π/2 ≤ x ≤ π), it becomes negative.
Since -π/2 ≤ x ≤ 0, we are working in the fourth quadrant. In the fourth quadrant, the cosine value is positive, and the corresponding reference angle in the first quadrant is θ = π - θ'. We can find θ' using the Pythagorean theorem.
Using the triangle we drew earlier, the opposite side can be found by subtracting the adjacent side from the hypotenuse. Therefore, the opposite side is -1.
Now, using the Pythagorean theorem, we have:
opposite^2 + adjacent^2 = hypotenuse^2
(-1)^2 + (√3)^2 = 2^2
1 + 3 = 4
4 = 4
Since this equation is true, we have a valid right triangle.
Now, let's find the angle θ' in the first quadrant using the reference angle:
sin(θ') = opposite/hypotenuse = -1/2
We know that sin(π/6) = 1/2, so θ' = π/6.
Finally, to find the angle x in the fourth quadrant, we subtract θ' from π:
x = π - θ' = π - π/6 = 5π/6.
Therefore, the exact value of x, in radians, is 5π/6.