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Question
Find the trigonometric functions of :
−240°
Solution
Trigonometric Functions of ( – 240°) :
Let the measure of ∠XOA in standard position be − 240°.
Its terminal arm (ray OA) intersects the x· standard unit circle in P (x, y), which lies in the second quadrant.
Draw segment PM perpendicular to the X-axis.
Then OM =l x l and MP = l y l.
In right-angled triangle OMP,
m∠MOP = 60° and OP= 1
∴ m∠MPO = 30°
∴ OM = `1/2"OP" = 1/2 xx 1 = 1/2`
∴ y2 = MP2 = 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 second quadrant.
∴ x < 0 and y > 0
∴ x = `-1/2 and y = sqrt(3)/2`
∴ P is `(-1/2, sqrt(3)/2)`
∴ sin (− 240°) = y = `sqrt(3)/2`
cos (− 240°) = x = `-1/2`
tan (− 240°) = `y/x = ((sqrt(3)/2))/((-(1)/2)) = -sqrt(3)`
cosec (− 240°) = `1/y = 1/((sqrt(3)/2)) = 2/sqrt(3)`
sec (− 240°) = `1/x = 1/((-1/2))` = − 2
cot (− 240°) = `x/y = ((-(1)/2))/((sqrt(3)/2)) = -1/sqrt(3)`.
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