Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 372: 16

$\sin(y)=\sqrt {1-x^{2}}$

Using Pythagorouses's theorem, the third side, $k$, of this right angled triangle can be found: $1^{2}=x^{2}+ k^{2}$ $k^{2}=1-x^{2}$ $k=\sqrt {1-x^{2}}$ Using the triangle given in the diagram, the equation for $\sin(y)=\frac{opposite}{hypotenuse}$ opposite $=\sqrt {1-x^{2}}$, hypotenuse $=1$ Therefore $\sin(y)=\frac{\sqrt {1-x^{2}}}{1}$ $=\sqrt {1-x^{2}}$