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प्रश्न
Find the trigonometric functions of :
−30°
उत्तर
Trigonometric Functions of ( − 30°) :
Let measure of ∠XOA in standard position be − 30°.
Its terminal arm (ray OA) intersects the standard unit circle in P (x, y), which lies in the fourth 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 = 30° and OP = 1.
∴ MP = `1/2"OP" = 1/2 xx 1 = 1/2`.
∴ x2 = OM2 = OP2 – MP2
= `1^2 - (1/2)^2 = 1 - 1/4 = 3/4`
∴ x = `±sqrt(3)/2 and y = ±1/2`
But P lies in the fourth quadrant.
∴ x > 0 and y < 0
∴ x = `sqrt(3)/2 and y = -1/2`
∴ P is `(sqrt(3)/2, -1/2)```
∴ sin (− 30°) = y = `-1/2`
cos (− 30°) = x = `sqrt(3)/2`
tan (− 30°) = `y/x = ((-1/2))/(((sqrt(3))/2)) = -1/sqrt(3)`
cosec (− 30°) = `1/y = 1/((-(1)/2))` = – 2
sec ( − 30°) = `1/x = 1/((sqrt(3)/2)) = 2/sqrt(3)`
cot (− 30°) = `x/y = ((sqrt(3)/2))/((-(1)/2)) = -sqrt(3)`
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