Trigonometry (11th Edition) Clone

$$\sin\theta=\frac{\sqrt5}{5}$$
$$\cos2\theta=\frac{3}{5}\hspace{1cm}\theta\hspace{0.2cm}\text{terminates in quadrant I}\hspace{1cm}\sin\theta=?$$ From the original half-angle identity for sines: $$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$ we can rewrite to use to find $\sin\theta$ $$\sin\theta=\pm\sqrt{\frac{1-\cos2\theta}{2}}$$ Also, whether to choose positive or negative square root now depends on the sign of $\sin\theta$. $\theta$ terminates in quadrant I, where sines are positive. Therefore, $\sin\theta\gt0$. As a result, we need to choose positive square root: $$\sin\theta=\sqrt{\frac{1-\cos2\theta}{2}}$$ Now we can start calculating $\sin\theta$. $$\sin\theta=\sqrt{\frac{1-\frac{3}{5}}{2}}$$ $$\sin\theta=\sqrt{\frac{\frac{2}{5}}{2}}$$ $$\sin\theta=\sqrt{\frac{2}{5\times2}}=\sqrt{\frac{1}{5}}$$ $$\sin\theta=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}$$ Therefore, $$\sin\theta=\frac{\sqrt5}{5}$$