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Find the trigonometric functions of : −315° - Mathematics and Statistics

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प्रश्न

Find the trigonometric functions of :

−315°

योग

उत्तर

Trigonometric Functions of (−315°) :


Let m∠XOA = −315°

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

Draw seg PM perpendicular to the X-axis.

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

∴ ΔOMP is a 45° – 45° – 90° triangle.

m∠MOP = 45° and OP = 1.

∴ OM = MP

∴ | x | = | y |

By the distance formula,

x2 + y2 = 1

∴ x2 + x2 = 1

∴ 2x2 = 1

∴ x2 = `1/2`

∴ x = `± 1/sqrt2`

∴ `x = ± 1/sqrt2, y = ± 1/sqrt2`

Since point P lies in the 1st quadrant,

x > 0, y > 0

∴ x = OM = `1/sqrt(2) and y = "MP" = 1/sqrt(2)`

∴ P ≡ `(1/sqrt(2), 1/sqrt(2))`

sin (−315°) = y = `1/sqrt(2)`

cos (−315°) = x = `1/sqrt(2)`

tan (−315°) = `y/x = (1/sqrt(2))/(1/sqrt(2))` = 1

cosec (−315°) = `1/y = 1/((1/sqrt(2))) = sqrt(2)`

sec (−315°) = `1/x = 1/((1/sqrt(2))) = sqrt(2)`

cot (−315°) = `x/y = (1/sqrt(2))/(1/sqrt(2))` = 1

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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) xviii) | पृष्ठ २१
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