# Finding the Exact Value of cos θ?

Here's the given question: I know that if I drew out the Cartesian plane, it'll form a right triangle in QII (that means cos θ must be negative).

My plan was to use $r=\sqrt{x^2 + y^2}$ (aka the Pythagorean Theorem), since, by definition of soh cah toa, I'm given the opposite (the leg parallel to the y-axis, $\sqrt{5}$) and hypotenuse, $5$.

$5=\sqrt{x^2 + (\sqrt{5})^2}$

$5= x + \sqrt{5}$

$5 -\sqrt{5} = x$

However, I know that this is wrong since the answer cannot be positive since it's in QII... plus it's none of the options... so what did I do wrong?

## 4 Answers

First, you've done something bad that you should be aware of: $$\sqrt{a^2+b^2}\neq a+b,$$ so your second line of calculation is not correct.

Now, to find $x$, we should start as you did with $$5=\sqrt{x^2+(\sqrt 5)^2}.$$ Continuing, we get \begin{align*} 5^2 &= x^2 +5\\ 20 &= x^2\\ \sqrt 20 = x. \end{align*}

Thus $\cos\theta$ is the (negative!) ratio $-\frac{x}{5}=-\frac{\sqrt 20}{5}=-\frac{2\sqrt 5}{5}$, so the answer is (D).

• 90<x<180 !!! – Rhys Hughes Jun 10 '18 at 21:14

HINT

Recall that

$$\sin^2 \theta + \cos^2 \theta =1$$

then

$$\cos^2 \theta = 1-\sin^2 \theta = 1-\frac15=\frac45 \implies \cos \theta=\pm\frac2{\sqrt 5}=\pm\frac{2\sqrt 5}5$$

finally recall that for $\theta \in [\pi/2,\pi]$ we have $\cos \theta \le 0$.

Hint: You've fallen victim to Freshman's dream, a common mistake associated with exponentiation. Specifically, your algebra mistake lies in the first deduction:

$$5 = \sqrt{x^2 + \sqrt{5}^2}$$ $$\color{red}{5 = x + \sqrt 5}$$

In general:

$$\sqrt{a^2 + b^2} \neq a + b$$

Instead, square both sides and solve:

$$25 = x^2 + 5 \implies x^2 = 20 \implies x = \pm 2\sqrt 5$$

Then find $\cos \theta$ with these new $x$ values (make sure to pick the right $x$ value that corresponds to the correct quadrant).

In the range $90^0\le x\le180^0, -1\le \cos(x) \le 0$.

Take a look at this triangle: Note that via Pythagoras, $c=\sqrt{a^2+b^2}$.

Then basic trigonometry tells us: $\sin\theta=\frac {b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\frac {a}{\sqrt{a^2+b^2}}$

We know $\sin\theta=\frac{\sqrt{5}}{5}$, thus $b=\sqrt{5}, c=5$ and we can use that: $$5=\sqrt{\sqrt{5}^2+a^2}\to a^2=20\to a=2\sqrt5$$

From this, $\cos\theta=\frac{2\sqrt5}{5}$, but since we know it's less than $0$, it is simply the negative, so $\cos\theta=-\frac{2\sqrt5}{5}$