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प्रश्न
Find the trigonometric functions of :
−45°
उत्तर
Angle of measure (– 45°):
Let m∠XOA = −45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle
OP = 1,
OM = `1/sqrt(2)"OP"`
= `1/sqrt(2)(1)`
= `1/sqrt(2)`
PM = `1/sqrt(2)"OP"`
= `1/sqrt(2)(1)`
= `1/sqrt(2)`
Since point P lies in the 4th quadrant,
x > 0, y < 0
∴ x = OM = `1/sqrt(2)` and y = - PM = `(-1)/sqrt(2)`
∴ P ≡ `(1/sqrt(2), (-1)/sqrt(2))`
sin (– 45°) = y = `- 1/sqrt(2)`
cos (– 45°) = x = `1/sqrt(2)`
tan (– 45°) = `y/x = ((-1/sqrt(2)))/((1/sqrt(2)))` = – 1
cosec (– 45°) = `1/y = 1/((-1/sqrt(2))) = -sqrt(2)`
sec (– 45°) = `1/x = 1/((1/sqrt(2))) = sqrt(2)`
cot (– 45°) = `x/y = ((1/sqrt(2)))/((-1/sqrt(2))` = – 1
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