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Find sin(2x), cos(2x), and tan(2x) if cos(x) = -8/17 and x terminates in quadrant III.

a) sin(2x) = -15/17, cos(2x) = -64/289, tan(2x) = 15/64
b) sin(2x) = -15/17, cos(2x) = 64/289, tan(2x) = -15/64
c) sin(2x) = 15/17, cos(2x) = 64/289, tan(2x) = -15/64
d) sin(2x) = 15/17, cos(2x) = -64/289, tan(2x) = 15/64

1 Answer

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Final answer:

To find sin(2x), cos(2x), and tan(2x) with cos(x) = -8/17 in the third quadrant, we calculate sin(x) using the Pythagorean identity, then use double-angle formulas to find the values. The correct option is A.

Step-by-step explanation:

To find sin(2x), cos(2x), and tan(2x) given that cos(x) = -8/17 and x terminates in quadrant III, we need to utilize the double-angle formulas for sine and cosine:

  • sin(2x) = 2sin(x)cos(x)
  • cos(2x) = cos²(x) - sin²(x)
  • tan(2x) = sin(2x) / cos(2x)

Given cos(x) = -8/17, and knowing that we are in quadrant III where sine is also negative, we can find sin(x) using the Pythagorean identity:

sin²(x) = 1 - cos²(x) -> sin(x) = √(1 - cos²(x)) = √(1 - (-8/17)²) -> sin(x) = -√(289 - 64)/17 -> sin(x) = -15/17

Now we can find the requested functions:

  • sin(2x) = 2 * (-15/17) * (-8/17) = 240/289
  • cos(2x) = (-8/17)² - (-15/17)² = (64/289) - (225/289) = -161/289
  • tan(2x) = sin(2x) / cos(2x) = (240/289) / (-161/289) = -240/161

The correct option matching these values is option A: sin(2x) = -15/17, cos(2x) = -64/289, tan(2x) = 15/64.

User Mansinh
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