Six Trigonometric Functions

Question

If tan(θ)=−√3, find all the angles which satisfy the given equation and condition.

Answer 1

tanx = -sqr3, that's quadrant II or IV, the only quadrants where tangents are negative

the angle is 2pi/3 + npi where n is any integer

the reference angle is 60 degrees or pi/3. It's a 1-sqr3- 2 right triangle

Answer 2

tan(theta) = -square root(3) at theta = 120 and 300 (where sin and cos have opposite signs)

The reference angle (angle made with horizontal axis) for 120 and 300 degrees is 60 degrees.

120 is in QII where sin is positive and cos is negative

60 is refence angle - compute trig functions as if angle was 60 then apply correct sign

sin(120) = sqrt root(3)/2

cos(120) = -1/2

cot(120) = 1/tan(120) = 1/-sqrt(3) = -sqrt(3)/3

csc(120) = 1/sin(120) = 2/sqrt root(3) = 2*sqrt(3)/3

sec(120) = 1/cos(120) = -2

300 is in QIV where sin is negative and cos is positive

60 is reference angle - compute trig functions as if angle was 60 then apply correct sign

sin(300) = -sqrt root(3)/2

cos(300) = 1/2

cot(300) = 1/tan(300) = 1/-sqrt(3) = -sqrt(3)/3

csc(300) = 1/sin(300) = -2/sqrt root(3) = -2*sqrt(3)/3

sec(300) = 1/cos(300) = 2