Given sine, theta, equals, start fraction, 3, divided by, 4, end fractionsinθ=
4
3
and angle thetaθ is in Quadrant I, what is the exact value of cosine, thetacosθ in simplest form? Simplify all radicals if needed.
Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we can solve for $\cos(\theta)$.
Since $\sin(\theta) = \frac{4}{3}$, we can substitute this value into the identity:
$\left(\frac{4}{3}\right)^2 + \cos^2(\theta) = 1$
Simplifying, we have:
$\frac{16}{9} + \cos^2(\theta) = 1$
Subtracting $\frac{16}{9}$ from both sides:
$\cos^2(\theta) = 1 - \frac{16}{9} = \frac{9}{9} - \frac{16}{9} = - \frac{7}{9}$
Since $\theta$ is in Quadrant I, $\cos(\theta)$ is positive. Taking the square root of both sides:
$\cos(\theta) = \sqrt{- \frac{7}{9}}$
However, since $\cos(\theta)$ must be positive, the simplified value of $\cos(\theta)$ is:
$\cos(\theta) = \frac{\sqrt{7}}{\sqrt{9}} = \frac{\sqrt{7}}{3}$