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3 February, 01:59

If cos theta = 3/5 and theta is in quadrant 4, sin 2 theta =

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  1. 3 February, 02:29
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    Hello!

    In the first quadrant, all the trigonometric functions are positive.

    In the second quadrant, sin and csc are positive, the rest are negative.

    In the third quadrant, tan and cot are positive; the rest are negative.

    In the fourth quadrant, cos and sec are positive, and the rest are negative.

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    To find sin 2θ, we use the equation: sin 2θ = 2 sinθ cosθ.

    Since we are given cos θ, which is 3/5, we need to find sin θ.

    cos θ = x / r

    sin θ = y / r

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    To find y, we need to use the Pythagorean Theorem.

    3² + y² = 5²

    9 + y² = 25 (subtract 9 from both sides)

    y² = 16 (take the square root of both sides)

    y = 4

    Since sin θ is negative in quadrant 4, the y-value of sin θ is negative.

    Therefore, sin θ = - 4/5.

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    Then, we can substitute these values into sin 2θ = 2sinθ cosθ

    sin 2θ = 2 (-4/5) (3/5)

    sin 2θ = (-8/5) (3/5)

    sin 2θ = - 24/25

    Therefore, sin 2 theta is equal to - 24/25.
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