If cos x=−4/5, and π<x<3π2, what is sin(x+π/2)?
i really need help...
You sure do
perhaps you mean
If cos x=−4/5, and π<x<3π / 2, what is sin(x+π/2)?
so x is in quadrant 3, this is a 3,4, 5 right triangle with x = -4, y = -3
keep going counterclockwise 90 degrees (π/2 radians) more
You are now in quadrant 4 with x = 3 and y = -4
the sin is then -4/5
sin(x+π/2) = sinx cosπ/2 + cosx sinπ/2 = cosx
thank u!
hey obleck, how doi i use that in this:
If tan x=√3, and 180°<x<270°, what is sin(x−210°)?
or any other user...
To find the value of sin(x+π/2), we'll first determine sin(x) using the given information: cos(x) = -4/5.
We know that sin^2(x) + cos^2(x) = 1.
Since we have cos(x) = -4/5, we can find sin(x) using this equation:
sin^2(x) + (-4/5)^2 = 1
sin^2(x) + 16/25 = 1
sin^2(x) = 1 - 16/25
sin^2(x) = 9/25
Taking the square root of both sides, we have:
sin(x) = ± √(9/25)
sin(x) = ± 3/5
Since π < x < 3π/2, it means that x is within the third quadrant. In the third quadrant, both sin(x) and cos(x) are negative. Therefore, sin(x) = -3/5.
Now, to find sin(x+π/2), we need to substitute the value of sin(x) into the formula for sin(a + b):
sin(a + b) = sin(a) × cos(b) + cos(a) × sin(b)
In this case, a = x and b = π/2:
sin(x + π/2) = sin(x) × cos(π/2) + cos(x) × sin(π/2)
cos(π/2) = 0 and sin(π/2) = 1; substituting these values and sin(x) = -3/5:
sin(x + π/2) = (-3/5) × 0 + (-4/5) × 1
sin(x + π/2) = 0 + (-4/5)
sin(x + π/2) = -4/5
Therefore, sin(x + π/2) equals -4/5.