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Given that sin(θ) = -24/25, and θ is in quadrant iii, what is cos(2θ)?

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User Corlettk
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1 Answer

4 votes

Final answer:

To find cos(2\(\theta\)) given sin(\(\theta\)) = -24/25 and \(\theta\) in quadrant III, we use the double-angle formula and properties of trigonometric functions in the different quadrants to calculate it as -527/625.

Step-by-step explanation:

If we have that sin(\(\theta\)) = -24/25 and \(\theta\) is in quadrant III, we can find cos(2\(\theta\)) by using the double-angle formula for cosine, which can be stated as:

  • \(cos(2\theta) = cos^2(\theta) - sin^2(\theta)\)
  • \(cos(2\theta) = 2cos^2(\theta) - 1\)
  • \(cos(2\theta) = 1 - 2sin^2(\theta)\)

Since \(sin(\theta) = -24/25\), we can find \(sin^2(\theta)\), which is \((24/25)^2 = 576/625\). Now, because \(\theta\) is in the third quadrant, the cosine of \(\theta\) must be negative (since in the third quadrant, both sine and cosine are negative). So, we can say that \(cos(\theta) = -\sqrt{1 - sin^2(\theta)} = -\sqrt{1 - 576/625} = -\sqrt{49/625} = -7/25\).

Now using the double-angle formula \(cos(2\theta) = 1 - 2sin^2(\theta)\), we get:

\(cos(2\theta) = 1 - 2*(576/625) = 1 - 1152/625 = (625 - 1152)/625 = -527/625\).

The value of cos(2\(\theta\)) is -527/625.

answered
User Gkd
by
8.6k points
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