1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Because $\sin^2 x + \cos^2 x = 1$ and $\sin x / \cos x = 2$ gives enough information to solve. $\endgroup$ – user296602 Jan 30 '18 at 23:20
  • 1
    $\begingroup$ $x$ 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. $\endgroup$ – orole Jan 30 '18 at 23:29
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I can totally see it now, but I would have never thought of that! thanks $\endgroup$ – Pablo Jan 30 '18 at 23:27
0
$\begingroup$

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

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.