Don't understand implication $\tan(t)=2 \implies x=\pm\frac{1}{\sqrt{5}} \text{and}\ y=\pm\frac{2}{\sqrt{5}}.$

Question

Let $x=\cos(t)$ and $y=\sin(t)$. $$\frac{\sin(t)}{\cos(t)}=\tan(t)=2 \implies x=\pm\frac{1}{\sqrt{5}} \text{and}\ y=\pm\frac{2}{\sqrt{5}}.$$

I don't understand how the textbook arrived at the above implication?

Answer 1

$$[\sin^2 x + \cos ^2 x = 1]~\text{and}~[\sin(x) = 2\cos(x)] \Rightarrow 5\cos ^2 x = 1$$ $$\cos ^2 x = \frac15\Rightarrow \cos x = \pm\frac15\Rightarrow\sin x = \pm\frac25$$

Answer 2

$$\implies\dfrac{\sin t}2=\dfrac{\cos t}1=\pm\sqrt{\dfrac{\sin^2t+\cos^2t}{2^2+1^2}}$$

Answer 3

Let $\;\Delta ABC\;$ be a right triangle, with $\;\angle B=90^\circ\;$, and with lengths (using Pythagoras Theorem):

$$AB=2x\;,\;\;BC=x\;,\;\;AC=\sqrt5\,x$$

In this triangle we get:

$$t:=\angle BCA\implies\tan t=\frac{AB}{BC}=2\;,\;\;\sin t=\frac{AB}{AC}=\frac2{\sqrt5}\;,\;\;\cos t=\frac{BC}{AC}=\frac1{\sqrt5}$$

and now check, on the trigonometric circle, the different mixes of signs...