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प्रश्न
Find the trigonometric functions of :
−315°
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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