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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

find sin 2x, cos 2x, tan 2x given that cos x = 4/5, 3pi/2<x<2pi?

i have to show work too!!! 

3 Answers

Relevance
  • 1 month ago

    cos²(x) + sin²(x) = 1

    sin²(x) = 1 - cos²(x) → given that: cos(x) = 4/5

    sin²(x) = 1 - (4/5)²

    sin²(x) = (25/25) - (16/25)

    sin²(x) = 9/25

    sin(x) = ± 3/5 → given that: 3π/2 < x < 2π → sin(x) < 0

    sin(x) = - 3/5

    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)

    sin(2x) = 2.(- 3/5).(4/5)

    sin(2x) = - 24/25

    Do you know this identity?

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

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

    cos(2x) = cos²(x) - sin²(x)

    cos(2x) = (4/5)² - (9/25)

    cos(2x) = (16/25) - (9/25)

    cos(2x) = 7/25

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

    tan(2x) = (- 24/25)/(7/25)

    tan(2x) = - 24/7

  • Ash
    Lv 7
    1 month ago

    sin²x + cos²x = 1

    sin²x + (⅘)² = 1

    sin²x + 16/25 = 1

    sin²x = 1 - 16/25

    sin²x = 9/25

    sinx = ±3/5 

    Given 3𝜋/2 < x < 2𝜋 means x is in 4th quadrant

    In that quadrant sin is negative.

    so we get 

    sinx = -⅗

    sin2x = 2 sinx cosx

    sin2x = 2(-⅗)(⅘)

    sin2x = -24/25

    cos2x = 2cos²x - 1

    cos2x = 2(⅘)² - 1

    cos2x = 7/25

    tan2x = sin2x / cos2x

    tan2x = (-24/25)/(7/25)

    tan2x = -24/7

     

  • ?
    Lv 7
    1 month ago

    Hint: use your trig identities. Do you know what these are?

    Hopefully no one will spoil you the answers. That would be very irresponsible of them.

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