## Algebra and Trigonometry 10th Edition

$\arccos \dfrac{x}{6}=\arcsin \dfrac{\sqrt {36-x^2}}{6}$
Consider $\theta=\arcsin \dfrac{\sqrt {36-x^2}}{6}$ We can see that $\sqrt {36-x^2}$ is a positive quanity and $\theta$ must lie in the first quadrant. So, we have: $\cos \theta=\dfrac{x}{6}$ and $\theta =\arccos \dfrac{x}{6}$ and $\arccos \dfrac{x}{6}=\arcsin \dfrac{\sqrt {36-x^2}}{6}$