Question

Answer the following:

Find the trigonometric functions of :

−300°

Answer 1

Trigonometric Functions of ( − 300°) :

Let the measure of ∠XOA in standard position be (– 300°).

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

Draw the segment PM perpendicular to the X-axis.

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

In right-angled triangle OMP,

m∠MOP = 60° and OP = 1

∴ m∠MPO = 30°

∴ OM = `1/2"OP" = 1/2 xx 1 = 1/2`

∴ | x | = `1/2`

By the distance formula,

x² + y² = 1

∴ `(1/2)^2 + y^2` = 1

∴ `1/4 + y^2` = 1

∴ y² = `1 - 1/4 = 3/4`

∴ y = `± sqrt(3)/2 and x = ±1/2`

But P lies in the first quadrant.

∴ x > 0 and y > 0

∴ x = `1/2 and y = sqrt(3)/2`

∴ P is `(1/2, sqrt(3)/2)`

∴ sin (– 300°) = y = `sqrt(3)/2`

cos (– 300°) = x = `1/2`

tan (– 300°) = `y/x = ((sqrt(3)/2))/((1/2)) = sqrt(3)`

cosec (– 300°) = `1/y = 1/((sqrt(3)/2)) = 2/sqrt(3)`

sec (– 300°) = `1/x = 1/((1/2))` = 2

cot (– 300°) = `x/y = ((1/2))/((sqrt(3)/2)) = 1/sqrt(3)`