Solve: 3tan^2x-1=0. I got +/- (the sq. root of 3)/3 which is correct according to my study guide. However, I don't know how to use the unit circle to find this. The copy of the unit circle which I have does not have tangent values listed, only sine and cosine. I've tried googling unit circle with no avail. Would you be able to share with me what the tan values are for pi/6, pi/4 and pi/3? Thanks so much.
tan(x) = sin(x) / cos(x)
Of course! I'd be happy to help you with the values of tangent (tan) for π/6, π/4, and π/3.
To find the values of tangent on the unit circle, you can use the following formula:
tanθ = sinθ / cosθ
Let's find the values step by step:
1. Start by finding the values of sine (sin) and cosine (cos) for the given angles on the unit circle:
For π/6 (30 degrees):
sin(π/6) = 1/2
cos(π/6) = √3/2
For π/4 (45 degrees):
sin(π/4) = √2/2
cos(π/4) = √2/2
For π/3 (60 degrees):
sin(π/3) = √3/2
cos(π/3) = 1/2
2. Now, substitute these values into the formula for tangent:
tan(π/6) = sin(π/6) / cos(π/6) = (1/2) / (√3/2) = √3/3
tan(π/4) = sin(π/4) / cos(π/4) = (√2/2) / (√2/2) = 1
tan(π/3) = sin(π/3) / cos(π/3) = (√3/2) / (1/2) = √3
So, the values of tangent for π/6, π/4, and π/3 are:
tan(π/6) = √3/3
tan(π/4) = 1
tan(π/3) = √3
You can use these values to verify your solution to the equation 3tan^2x - 1 = 0 and continue with your calculations.