Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 99

See graph. a. $y=2sin(x)+1$ b. see explanations.

Graph the equation as shown in the figure. a. The graph appears to be the same as $y=2sin(x)+1$ b. Using the identity $cos(2x)=1-2sin^2x$, we have $y=\frac{1-2(1-2sin^2x)}{2sin(x)-1}=\frac{4sin^2x-1)}{2sin(x)-1}=\frac{(2sin(x)+1)(2sin(x)-1)}{2sin(x)-1}=2sin(x)+1$ provided that $2sin(x)-1\ne 0$