What's the value of sin(2x) if tan x = 0.8 and sin>0?

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  • 3 years ago
    Favorite Answer

    tan(x) = 0.8 → where: sin(x) > 0

    You can see that tan(x) > 0, so the angle x belongs to the quadrant I or quadrant III.

    Given that sin(x) > 0, you can deduce that the angle x is located in the quadrant I.

    tan(x) = 0.8

    sin(x)/cos(x) = 0.8

    sin(x) = 0.8 * cos(x)

    cos²(x) + sin²(x) = 1

    cos²(x) + [0.8 * cos(x)]² = 1

    cos²(x) + [0.8² * cos²(x)] = 1

    cos²(x).[1 + 0.8²] = 1

    cos²(x) * 1.64 = 1

    cos²(x) = 1/1.64

    cos²(x) = (± 1/√1.64)² → given that x is located in the quadrant I, you can say that: cos(x) > 0

    cos(x) = 1/(√1.64) ← memorize this result as (1)

    cos²(x) + sin²(x) = 1

    sin²(x) = 1 - cos²(x)

    sin²(x) = 1 - (1/1.64)

    sin²(x) = 0.64/1.64

    sin²(x) = [± √(0.64/1.64)]²

    sin²(x) = [± 0.8/√(1.64)]² → given that x is located in the quadrant I, you can say that: sin(x) > 0

    sin(x) = 0.8/(√1.64) ← memorize this result as (2)

    Do you know this identity?

    sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = b = x

    sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

    sin(2x) = 2.sin(x).cos(x) → recall (2): sin(x) = 0.8/(√1.64)

    sin(2x) = 2.[0.8/(√1.64)].cos(x) → recall (1): cos(x) = 1/(√1.64)

    sin(2x) = 2.[0.8/(√1.64)].[1/(√1.64)]

    sin(2x) = (2 * 0.8) / (√1.64 * √1.64)

    sin(2x) = 1.6/1.64

  • 3 years ago

    sin(2x)>0=> 2x is in the 1st or the 2nd quadrant.

    tan x=0.8=>x=38.6598082...=>2x=77.3196165...

    Thus sin(2x)=sin(77.3196165..)=0.97561 approximately.

  • Como
    Lv 7
    3 years ago

    tan x = 4/5

    sin x = 3 / √41

    cos x = 5 / √41

    sin 2x = 2 sin x cos x

    sin 2x = 2 [ 3 / √41 ] [ 5 / √41 ]

    sin 2x = 30 / 41

  • TomV
    Lv 7
    3 years ago

    sin(2x) = 2sin(x)cos(x)

    If sin(x) > 0 and tan(x) > 0, then x is in Q1

    tan(x) = sin(x)/cos(x) = 0.8

    sin(x) = 0.8cos(x)

    sin(2x) = 2(0.8cos²(x)) = 1.6cos²(x)

    sin²(x) = 0.64cos²(x) = 1 - cos²(x)

    1.64cos²(x) = 1

    cos²(x) = 1/1.64

    sin(2x) = 1.6/1.64 ≈ 0.97561

    =========================

    Corroboration:

    [tan(x) = 0.8 → x = arctan(0.8) ≈ 38.660°

    sin(2x) ≈ sin(2*38.660) ≈ sin(77.330) ≈ 0.97561]

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  • Mike G
    Lv 7
    3 years ago

    Use the tan half-angle formula

    sin(2x) = 2tanx/[1+tan^2(x)] = 1.6/1.64

    = 40/41

  • 3 years ago

    To keep it simple, do things one easy step at a time, and use the basic identitied.

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

    therefore, cos^2(x) = 1 - sin^2(x)

    and

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

    tan(x) = 0.8

    square both sides

    tan^2(x) = 0.64

    tan^2(x) = sin^2(x) / cos^2(x)

    tan^2(x) = sin^2(x) / (1 - sin^2(x)) = 0.64

    solve for sin^2(x)

    sin^2(x) = 0.64(1 - sin^2(x))

    sin^2(x) = 0.64 - 0.64*sin^2(x)

    1.64 sin^2(x) = 0.64

    sin^2(x) = 0.64 / 1.64 = 0.390244 (approx.)

    square root both sides

    sin(x) = 0.624695

    set that aside for now

    sin^2(x) = 0.390244

    therefore

    cos^2(x) = 1 - 0.390244 = 0.609756

    cos(x) = 0.3718025

    sin(2x) = 2sin(x)cos(x) = 0.464526

    ---

    Yes, there are shortcuts. But they require you to memorize more complicated things.

  • alex
    Lv 7
    3 years ago

    tan x = 0.8 and sin>0 ---> cosx = 1/√1.64 and sinx = 0.8/√1.64

    sin(2x)=2sinx cosx=...

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