If csc^2 x=1.4 on the domain [0, π/2]. What is the value of tan x , rounded to the nearest tenth?

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  • 3 years ago
    Favorite Answer

    csc²(x) = 1.4 → you know that: csc(x) = 1/sin(x)

    [1/sin(x)]² = 1.4

    1/sin²(x) = 1.4

    sin²(x) = 1/1.4

    sin²(x) = (± 1/√1.4)²

    sin(x) = ± 1/√1.4 → on the domain [0 ; π/2], you can see that: sin(x) > 0

    sin(x) = 1/√1.4

    cos²(x) + sin²(x) = 1

    cos²(x) = 1 - sin²(x) → recall: sin²(x) = 1/1.4

    cos²(x) = 1 - (1/1.4)

    cos²(x) = (1.4/1.4) - (1/1.4)

    cos²(x) = (1.4 - 1)/1.4

    cos²(x) = 0.4/1.4

    cos²(x) = [± √(0.4/1.4)]²

    cos(x) = ± √(0.4/1.4) → on the domain [0 ; π/2], you can see that: cos(x) > 0

    cos(x) = √(0.4/1.4)

    cos(x) = (√0.4)/(√1.4)

    tan(x) = sin(x)/cos(x) → recall: sin(x) = 1/√1.4

    tan(x) = [1/(√1.4)]/cos(x) → recall: cos(x) = (√0.4)/(√1.4)

    tan(x) = [1/(√1.4)]/[(√0.4)/(√1.4)] → you can simplify by √1.4

    tan(x) = 1/√0.4

    tan(x) = √(1/0.4)

    tan(x) = √2.5

    tan(x) ≈ 1.5811388 → rounded to the nearest tenth

    tan(x) ≈ 1.6

  • Pope
    Lv 7
    3 years ago

    csc²(x) = 1.4

    csc²(x) - 1 = 0.4

    cot²(x) = 0.4

    1/cot²(x) = 2.5

    tan²(x) = 2.5

    tan(x) = √(2.5) ... positive because 0 < x < π/2

    tan(x) ≈ 1.6

  • 3 years ago

    csc(x) = 1/sin(x)

    sin(x) = 1/csc(x)

    sin^2(x) = 1/csc^2(x) = 1/1.4 = 1/(7/5) = 5/7

    sin(x) = +/- sqrt(5/7)

    The domain is only Quadrant I, so sin(x) must be positive, so sin(x) = sqrt(5/7)

    sin^2(x) + cos^2(x) = 1

    cos^2(x) = 1 - sin^2(x) = 1 - (5/7) = 2/7

    cos(x) = +/- sqrt(2/7)

    The domain is only Quadrant I, so cos(x) must be positive, so cos(x) = sqrt(2/7)

    tan(x) = sin(x)/cos(x)

    tan(x) = sqrt(5/7)/sqrt(2/7)

    tan(x) = sqrt((5/7)/(2/7))

    tan(x) = sqrt(5/2) = sqrt(2.5) =~ 1.5811388300841896659994467722164, round off to 1.6

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