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26 September, 22:17

If tan theta equals 15/8 in quadrant 3 and you need to find the sin, cos, and tan of double angle, in what quadrant would the double angle reside.

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  1. 27 September, 01:48
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    Answer: sin2θ = 240/289

    cos2θ = - 161/289

    tan2θ = - 240/161

    2θ is in quadrant 2

    Step-by-step explanation: tanθ = 15/8

    sinθ/cosθ = 15/8

    sinθ = 15cosθ/8

    sin²θ + cos²θ = 1

    (15cosθ/8) ² + cos²θ = 1

    225cos²θ/64 + cos²θ = 1

    225cos²θ + 64cos²θ = 64

    289cos²θ = 64

    cos²θ = 64/289

    cosθ = ±√64/289

    cosθ = ±8/17

    As tanθ in quad 3 ⇒ tanθ is +, cosθ is - and sinθ is -

    So, cosθ = - 8/17

    as sin θ = 15cosθ/8 = - 15/17

    For sin2θ = 2sinθcosθ = 2. (-15/17). (-8/17) = 240/289

    For cos2θ = cos²θ - sin²θ = (-8/17) ² - (-15/17) ² = - 161/289

    For tan2θ = sin2θ/cos2θ = (240/289) / (-161/289) = - 240/161

    As sin2θ + and cos2θ - ⇒ 2θ is in quadrant 2
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