Find tan0 if cos0= -1/2 and 0 terminates in quadrant II
In quadrant II
cos θ = - 1 / 2
for
θ = 2 π / 3 rad (120°)
sin θ = sin 2 π / 3
sin θ = √3 / 2
tan θ = sin θ / cos θ = ( √3 / 2 ) / ( - 1 / 2 ) = - √3
Cos A = x/r = -1/2
x^2 + y^2 = r^2
(-1)^2 + y^2 = 2^2
Y = sqrt3.
Tan A = y/x = sqrt3/-1 = -sqrt3.
To find tan(θ), we can use the relationship between sin(θ), cos(θ), and tan(θ) in a right triangle. Since we know that cos(θ) = -1/2 and θ terminates in the second quadrant, we can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to find sin(θ).
Step 1: Find sin(θ)
Since cos(θ) = -1/2, we can use the Pythagorean identity to find sin(θ):
sin^2(θ) + cos^2(θ) = 1
sin^2(θ) + (-1/2)^2 = 1
sin^2(θ) + 1/4 = 1
sin^2(θ) = 1 - 1/4
sin^2(θ) = 3/4
Taking the square root of both sides, we get:
sin(θ) = ±√(3/4)
Since θ terminates in the second quadrant, sin(θ) will be positive:
sin(θ) = √(3/4)
We can simplify this further:
sin(θ) = √(3)/2
Step 2: Find tan(θ)
We can now use the relationship between sin(θ) and cos(θ) to find tan(θ):
tan(θ) = sin(θ) / cos(θ)
tan(θ) = (√(3)/2) / (-1/2)
When dividing by a fraction, we multiply by its reciprocal:
tan(θ) = (√(3)/2) * (-2/1)
tan(θ) = -√(3)
Therefore, the value of tan(θ) is -√(3).