[Math] Finding other trigonometric ratios from one

algebra-precalculustrigonometry

I was given this answer:

enter image description here

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?

Best Answer

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

Right Triangle

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 $$\cos(\alpha)=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{1}{\sqrt{5}}$$

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