Precalculus Examples

$(- \frac{12}{13}, \frac{5}{13})$

Step 1

To find the $\cos (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(- \frac{12}{13}, \frac{5}{13})$ , draw the triangle between the three points $(0, 0)$ , $(- \frac{12}{13}, 0)$ , and $(- \frac{12}{13}, \frac{5}{13})$ .

Opposite : $\frac{5}{13}$

Adjacent : $- \frac{12}{13}$

Step 2

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

Tap for more steps...

Step 2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Tap for more steps...

Step 2.1.1

Apply the product rule to $- \frac{12}{13}$ .

$\sqrt{{(- 1)}^{2} {(\frac{12}{13})}^{2} + {(\frac{5}{13})}^{2}}$

Step 2.1.2

Apply the product rule to $\frac{12}{13}$ .

$\sqrt{{(- 1)}^{2} \frac{12^{2}}{13^{2}} + {(\frac{5}{13})}^{2}}$

Step 2.2

Raise $- 1$ to the power of $2$ .

$\sqrt{1 \frac{12^{2}}{13^{2}} + {(\frac{5}{13})}^{2}}$

Step 2.3

Multiply $\frac{12^{2}}{13^{2}}$ by $1$ .

$\sqrt{\frac{12^{2}}{13^{2}} + {(\frac{5}{13})}^{2}}$

Step 2.4

Raise $12$ to the power of $2$ .

$\sqrt{\frac{144}{13^{2}} + {(\frac{5}{13})}^{2}}$

Step 2.5

Raise $13$ to the power of $2$ .

$\sqrt{\frac{144}{169} + {(\frac{5}{13})}^{2}}$

Step 2.6

Apply the product rule to $\frac{5}{13}$ .

$\sqrt{\frac{144}{169} + \frac{5^{2}}{13^{2}}}$

Step 2.7

Raise $5$ to the power of $2$ .

$\sqrt{\frac{144}{169} + \frac{25}{13^{2}}}$

Step 2.8

Raise $13$ to the power of $2$ .

$\sqrt{\frac{144}{169} + \frac{25}{169}}$

Step 2.9

Combine the numerators over the common denominator.

$\sqrt{\frac{144 + 25}{169}}$

Step 2.10

Add $144$ and $25$ .

$\sqrt{\frac{169}{169}}$

Step 2.11

Divide $169$ by $169$ .

$\sqrt{1}$

Step 2.12

Any root of $1$ is $1$ .

$1$

Step 3

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{- \frac{12}{13}}{1}$ .

$\frac{- \frac{12}{13}}{1}$

Step 4

Divide $- \frac{12}{13}$ by $1$ .

$\cos (θ) = - \frac{12}{13}$

Step 5

Approximate the result.

$\cos (θ) = - \frac{12}{13} \approx - 0.\overline{923076}$