Question

Find the trigonometric functions of : 

−120°

Answer 1

Trigonometric Functions of (− 120°) :

Let the measure of ∠XOA in Y standard position be - 120°.

Its terminal arm (ray OA) intersects the standard unit circle in P (x, y), which lies in the third quadrant.

Draw seg PM perpendicular to the X-axis.

Then OM = |x| and MP = |y|.

In right-angled triangle OMP,

m∠MOP = 60° and OP = 1

∴ m∠MPO = 30°

∴ OM = `1/2"OP" = 1/2 (1) = 1/2`

∴ y² = PM²  = OP² - OM²

= `1^2 - (1/2)^2`

`= 1-1/4`

`= 3/4`

∴ y = `± sqrt(3)/2 and x = ± 1/2`

But P lies in the third quadrant.

∴ x < 0 and y < 0

∴ x = `-1/2 and y = -sqrt(3)/2`

∴ P is `(-1/2, -sqrt(3)/2)`

∴ sin (– 120°) = y = `-sqrt(3)/2`

cos (– 120°) = x = `-1/2`

tan (– 120°) = `y/x = ((-(sqrt(3))/2))/((-(1)/2)) = sqrt(3)`

cosec (– 120°) = `1/y = 1/((-(sqrt(3))/2)) = -2/sqrt(3)`

sec (– 120°) = `1/x = 1/((-(1)/2)` = – 2

cot (– 120°) = `x/y = ((-(1)/2))/((-sqrt(3)/2)) = 1/sqrt(3)`