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Find the trigonometric functions of : −30° - Mathematics and Statistics

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प्रश्न

Find the trigonometric functions of :

−30°

आकृति
योग

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 – MP

= `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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Trigonometric Functions of Specific Angles
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Trigonometry - 1 - EXERCISE 2.1 [पृष्ठ २१]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 2 Trigonometry - 1
EXERCISE 2.1 | Q 1) x) | पृष्ठ २१
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