Question

Question
Ana was asked to use triangles formed by angles on the unit circle to prove the Pythagorean Identity sin2θ+cos2θ=1, where sinθ=1617 and π2<θ<π. What value should she use for cosθ?(1 point)
Responses

−33√17
negative Start Fraction Start Root 33 End Root over 17 End Fraction

−117√
negative Start Fraction 1 over Start Root 17 End Root End Fraction

117√
Start Fraction 1 over Start Root 17 End Root End Fraction

33√17
Start Fraction Start Root 33 End Root over 17 End Fraction

Answer 1

To find the value of \( \cos \theta \) using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can start by substituting the given value of \( \sin \theta = \frac{16}{17} \).

1. Calculate \( \sin^2 \theta \):
\[
\sin^2 \theta = \left( \frac{16}{17} \right)^2 = \frac{256}{289}
\]

2. Use the Pythagorean identity to find \( \cos^2 \theta \):
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
\[
\frac{256}{289} + \cos^2 \theta = 1
\]

To isolate \( \cos^2 \theta \), subtract \( \frac{256}{289} \) from both sides:
\[
\cos^2 \theta = 1 - \frac{256}{289} = \frac{289}{289} - \frac{256}{289} = \frac{33}{289}
\]

3. Take the square root of both sides to find \( \cos \theta \):
\[
\cos \theta = \pm \sqrt{\frac{33}{289}} = \pm \frac{\sqrt{33}}{17}
\]

Since we are given that \( \pi/2 < \theta < \pi \) (which is the second quadrant), the cosine value must be negative in this range. Therefore, we take the negative value:
\[
\cos \theta = -\frac{\sqrt{33}}{17}
\]

The correct answer for \( \cos \theta \) is:
- negative Start Fraction Start Root 33 End Root over 17 End Fraction

So, the answer is:
\(-\frac{\sqrt{33}}{17}\)