We have the Pythagorean Identity:
\[ \cos^2\theta + \sin^2\theta = 1 \]
Substituting the given value of $\sin\theta$:
\[ \cos^2\theta + \left(\frac{7}{9}\right)^2 = 1 \]
\[ \cos^2\theta + \frac{49}{81} = 1 \]
Subtracting $\frac{49}{81}$ from both sides:
\[ \cos^2\theta = 1 - \frac{49}{81} \]
\[ \cos^2\theta = \frac{81}{81} - \frac{49}{81} \]
\[ \cos^2\theta = \frac{32}{81} \]
Taking the square root of both sides:
\[ \cos\theta = \pm \frac{\sqrt{32}}{\sqrt{81}} \]
\[ \cos\theta = \pm \frac{\sqrt{16 \cdot 2}}{9} \]
\[ \cos\theta = \pm \frac{\sqrt{16} \cdot \sqrt{2}}{9} \]
\[ \cos\theta = \pm \frac{4\sqrt{2}}{9} \]
Since $\frac{\pi}{2} < \theta < \pi$ and $\sin\theta > 0$, we know that $\cos\theta$ must be negative. Therefore, the correct answer is:
\[ \cos\theta = -\frac{4\sqrt{2}}{9} \]
So the answer is $\boxed{-\frac{4\sqrt{2}}{9}}$.
Using the Pythagorean Identity, determine cosθ if sinθ=7/9 and π/2<θ<π .(1 point) Responses −2√3 negative Start Fraction Start Root 2 End Root over 3 End Fraction 42√9 Start Fraction 4 Start Root 2 End Root over 9 End Fraction 2√3 Start Fraction Start Root 2 End Root over 3 End Fraction −42√9
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