Trigonometry Examples

$(\frac{\sqrt{13}}{7}, \frac{6}{7})$

Step 1

To find the $\cos (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{\sqrt{13}}{7}, \frac{6}{7})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{\sqrt{13}}{7}, 0)$ , and $(\frac{\sqrt{13}}{7}, \frac{6}{7})$ .

Opposite : $\frac{6}{7}$

Adjacent : $\frac{\sqrt{13}}{7}$

Step 2

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

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Step 2.1

Apply the product rule to $\frac{\sqrt{13}}{7}$ .

$\sqrt{\frac{{\sqrt{13}}^{2}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2

Rewrite ${\sqrt{13}}^{2}$ as $13$ .

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{13}$ as $13^{\frac{1}{2}}$ .

$\sqrt{\frac{{(13^{\frac{1}{2}})}^{2}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{\frac{13^{\frac{1}{2} \cdot 2}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2.3

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{\frac{13^{\frac{2}{2}}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2.4

Cancel the common factor of $2$ .

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Step 2.2.4.1

Cancel the common factor.

$\sqrt{\frac{13^{\frac{2}{2}}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2.4.2

Rewrite the expression.

$\sqrt{\frac{13^{1}}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.2.5

Evaluate the exponent.

$\sqrt{\frac{13}{7^{2}} + {(\frac{6}{7})}^{2}}$

Step 2.3

Simplify the expression.

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Step 2.3.1

Raise $7$ to the power of $2$ .

$\sqrt{\frac{13}{49} + {(\frac{6}{7})}^{2}}$

Step 2.3.2

Apply the product rule to $\frac{6}{7}$ .

$\sqrt{\frac{13}{49} + \frac{6^{2}}{7^{2}}}$

Step 2.3.3

Raise $6$ to the power of $2$ .

$\sqrt{\frac{13}{49} + \frac{36}{7^{2}}}$

Step 2.3.4

Raise $7$ to the power of $2$ .

$\sqrt{\frac{13}{49} + \frac{36}{49}}$

Step 2.3.5

Combine the numerators over the common denominator.

$\sqrt{\frac{13 + 36}{49}}$

Step 2.3.6

Add $13$ and $36$ .

$\sqrt{\frac{49}{49}}$

Step 2.3.7

Divide $49$ by $49$ .

$\sqrt{1}$

Step 2.3.8

Any root of $1$ is $1$ .

$1$

Step 3

$\cos (θ) = \frac{Adjacent}{Hypotenuse}$ therefore $\cos (θ) = \frac{\frac{\sqrt{13}}{7}}{1}$ .

$\frac{\frac{\sqrt{13}}{7}}{1}$

Step 4

Divide $\frac{\sqrt{13}}{7}$ by $1$ .

$\cos (θ) = \frac{\sqrt{13}}{7}$

Step 5

Approximate the result.

$\cos (θ) = \frac{\sqrt{13}}{7} \approx 0.51507875$