home > s25 ma2090 - dambroise precalculus > assessment\nhomework #12 - circular sine and cosine\nscore…

Explanation:

Step1: Use Pythagorean identity

We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. So, $\sin^{2}\theta=1 - \cos^{2}\theta$.

Step2: Substitute the given value of $\cos\theta$

Given $\cos\theta=\frac{5}{6}$, then $\sin^{2}\theta=1 - (\frac{5}{6})^{2}=1-\frac{25}{36}=\frac{36 - 25}{36}=\frac{11}{36}$.

Step3: Determine the sign of $\sin\theta$

Since $\theta$ is in the 1st - quadrant, $\sin\theta>0$. So, $\sin\theta=\sqrt{\frac{11}{36}}=\frac{\sqrt{11}}{6}$.

Answer:

$\frac{\sqrt{11}}{6}$