sin theta=-3/5 and theta is in the third quadrant, find the following: cos(theta+180/3)
since θ is in QIII, cosθ is also negative, and so is -4/5.
If you really meant what you wrote, then since 180/3 = 60, you want
cos(θ+60) = cosθ sinθ - cos60 sin60
= (-4/5)(-3/4) - (1/2)(√3/2)
= 3/5 - √3/4
If you meant cos (θ+180)/3 = cos(θ/3 + 60) then things are a bit harder
sinθ = -0.6 in QIII means θ=216.87
so, cos(θ/3 + 60) = cos(132.29) = -0.673
oops. misspoke on the first solution
cos(θ+60) = cosθ cos60 - sinθ sin60
= (-4/5)(√3/2) - (-3/5)(1/2)
= 3/10 - 2√3/5
To find cos(theta + 180/3), we need to determine the value of theta first.
Since sin(theta) = -3/5 and theta is in the third quadrant, we know that sin(theta) is negative.
First, let's find the value of cos(theta) using the given value of sin(theta) = -3/5.
We can use the Pythagorean identity to find cos(theta):
cos^2(theta) + sin^2(theta) = 1
Substituting sin(theta) = -3/5, we have:
cos^2(theta) + (-3/5)^2 = 1
cos^2(theta) + 9/25 = 1
cos^2(theta) = 1 - 9/25
cos^2(theta) = 16/25
Taking the square root of both sides, we get:
cos(theta) = ±4/5
Since theta is in the third quadrant and cos(theta) should be negative in the third quadrant, we have:
cos(theta) = -4/5
Now we can find cos(theta + 180/3). We add 180/3 to theta:
theta + 180/3 = theta + 60
Since theta is in the third quadrant, adding 60 degrees to it will bring us to the second quadrant. In the second quadrant, cos(theta) is negative.
Therefore, cos(theta + 180/3) is equal to -cos(θ + 60).
Now, we can use the trigonometric identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b), where a = θ and b = 60:
cos(θ + 60) = cos(θ)cos(60) - sin(θ)sin(60)
Substituting the values we found:
cos(θ + 60) = (-4/5)(1/2) - (-3/5)(√3/2)
cos(θ + 60) = -2/5 + 3√3/10
Therefore, cos(θ + 180/3) is equal to -2/5 + 3√3/10.