Calculate tan(3pi/2)
tan is found using Opposite/Adjacent
Since π < 270 < 3π/2 radians
it is located in Quadrant III
tan is positive.
270 > 90°, so it is obtuse
tan(3π/2) = N/A
θ° | θrad | sin(θ) | cos(θ) | tan(θ) | csc(θ) | sec(θ) | cot(θ) |
---|---|---|---|---|---|---|---|
0° | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
30° | π/6 | 1/2 | √3/2 | √3/3 | 2 | 2√3/3 | √3 |
45° | π/4 | √2/2 | √2/2 | 1 | √2 | √2 | 1 |
60° | π/3 | √3/2 | 1/2 | √3 | 2√3/3 | 2 | √3/3 |
90° | π/2 | 1 | 0 | N/A | 1 | 0 | N/A |
120° | 2π/3 | √3/2 | -1/2 | -√3 | 2√3/3 | -2 | -√3/3 |
135° | 3π/4 | √2/2 | -√2/2 | -1 | √2 | -√2 | -1 |
150° | 5π/6 | 1/2 | -√3/2 | -√3/3 | 2 | -2√3/3 | -√3 |
180° | π | 0 | -1 | 0 | 0 | -1 | N/A |
210° | 7π/6 | -1/2 | -√3/2 | √3/3 | -2 | -2√3/3 | √3 |
225° | 5π/4 | -√2/2 | -√2/2 | 1 | -√2 | -√2 | 1 |
240° | 4π/3 | -√3/2 | -1/2 | √3 | -2√3/3 | -2 | √3/3 |
270° | 3π/2 | -1 | 0 | N/A | -1 | 0 | N/A |
300° | 5π/3 | -√3/2 | 1/2 | -√3 | -2√3/3 | 2 | -√3/3 |
315° | 7π/4 | -√2/2 | √2/2 | -1 | -√2 | √2 | -1 |
330° | 11π/6 | -1/2 | √3/2 | -√3/3 | -2 | 2√3/3 | -√3 |