Finding other trigonometric ratios from one

So I was told that

$$\tan(x) = 2$$ Then, they said from this statement they could know that: $$\cos(x) = \frac{1}{\sqrt{5}}$$ $$\sin(x) = \frac{2}{\sqrt{5}}$$

Now, I understand that if I do

$$\tan^{-1}(2) = 63.4$$

And then after that I can get the ratio of cosine and sine. However, I don't know how they got the precise fraction. Does anyone know how?

• Because $\sin^2 x + \cos^2 x = 1$ and $\sin x / \cos x = 2$ gives enough information to solve. – user296602 Jan 30 '18 at 23:20
• $x$ is an angle in a right-angled triangle with catheti $2$ and $1$, such that you get $2=2/1=\tan(x)$. Then the hypotenuse is $\sqrt{2^2+1^2}=\sqrt{5}$, by Pythagoras. – orole Jan 30 '18 at 23:29

Recall that $\tan(\alpha)=\displaystyle\frac{\text{opposite}}{\text{adjacent}}=\dfrac{2}{1}$

By the Pythagorean theorem, the hypotenuse is $\sqrt{2^2+1^2}=\sqrt{5}$. Thus $$\sin(\alpha)=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{2}{\sqrt{5}}$$ and $$\cos(\alpha)=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{1}{\sqrt{5}}$$

You have $\tan (x)=\frac{\sin (x)}{\cos (x)}=c$ that implies $\sin (x) = c \cdot \cos (x)$. In your case $c=2$. Using the fondamental law of goniometry: $$\sin ^2(x)+\cos ^2(x)=1,$$ you have just to solve a system $$\begin{cases} \sin (x) = c \cdot \cos (x) \\ \sin ^2(x)+\cos ^2(x)=1 \end{cases}$$

In your case you have substituing the first equation to the second: $$5 \cos ^2 (x)=1$$ and so, if we consider the $\cos (x)$ be positive: $$\cos (x) = \frac{1}{\sqrt{5}}$$ Such that $\sin (x) = 2 \cos (x)$ we obtain: $$\sin (x) = 2 \cos (x) = \frac{2}{\sqrt{5}}$$ It's not an unique solution (if we don't have any angle constraints), because we could consider $\cos (x) = - \frac{1}{\sqrt{5}}$ and $\sin (x) = -\frac{2}{\sqrt{5}}$ too

• I can totally see it now, but I would have never thought of that! thanks – Pablo Jan 30 '18 at 23:27

$\tan x=2\implies \sin x=2\cos x$...

Now $\cos^2x+\sin^2x=1\implies \cos^2x+4\cos^2x=1\implies \cos^2x=\frac15\implies \cos x=\pm\frac{\sqrt5}5$.

Now $\sin x=\pm\sqrt{1-\cos^2x}=\pm\sqrt{1-\frac15}=\pm\frac{2\sqrt5}5$...

If we are in the first quadrant, take the positive values...