Trigonometry Examples

$(\frac{5}{8}, \frac{\sqrt{39}}{8})$

Step 1

To find the $\sin (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{5}{8}, \frac{\sqrt{39}}{8})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{5}{8}, 0)$ , and $(\frac{5}{8}, \frac{\sqrt{39}}{8})$ .

Opposite : $\frac{\sqrt{39}}{8}$

Adjacent : $\frac{5}{8}$

Step 2

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

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Step 2.1

Simplify the expression.

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Step 2.1.1

Apply the product rule to $\frac{5}{8}$ .

$\sqrt{\frac{5^{2}}{8^{2}} + {(\frac{\sqrt{39}}{8})}^{2}}$

Step 2.1.2

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

$\sqrt{\frac{25}{8^{2}} + {(\frac{\sqrt{39}}{8})}^{2}}$

Step 2.1.3

Raise $8$ to the power of $2$ .

$\sqrt{\frac{25}{64} + {(\frac{\sqrt{39}}{8})}^{2}}$

Step 2.1.4

Apply the product rule to $\frac{\sqrt{39}}{8}$ .

$\sqrt{\frac{25}{64} + \frac{{\sqrt{39}}^{2}}{8^{2}}}$

Step 2.2

Rewrite ${\sqrt{39}}^{2}$ as $39$ .

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{39}$ as $39^{\frac{1}{2}}$ .

$\sqrt{\frac{25}{64} + \frac{{(39^{\frac{1}{2}})}^{2}}{8^{2}}}$

Step 2.2.2

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

$\sqrt{\frac{25}{64} + \frac{39^{\frac{1}{2} \cdot 2}}{8^{2}}}$

Step 2.2.3

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

$\sqrt{\frac{25}{64} + \frac{39^{\frac{2}{2}}}{8^{2}}}$

Step 2.2.4

Cancel the common factor of $2$ .

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Step 2.2.4.1

Cancel the common factor.

$\sqrt{\frac{25}{64} + \frac{39^{\frac{2}{2}}}{8^{2}}}$

Step 2.2.4.2

Rewrite the expression.

$\sqrt{\frac{25}{64} + \frac{39^{1}}{8^{2}}}$

Step 2.2.5

Evaluate the exponent.

$\sqrt{\frac{25}{64} + \frac{39}{8^{2}}}$

Step 2.3

Simplify the expression.

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Step 2.3.1

Raise $8$ to the power of $2$ .

$\sqrt{\frac{25}{64} + \frac{39}{64}}$

Step 2.3.2

Combine the numerators over the common denominator.

$\sqrt{\frac{25 + 39}{64}}$

Step 2.3.3

Add $25$ and $39$ .

$\sqrt{\frac{64}{64}}$

Step 2.3.4

Divide $64$ by $64$ .

$\sqrt{1}$

Step 2.3.5

Any root of $1$ is $1$ .

$1$

Step 3

$\sin (θ) = \frac{Opposite}{Hypotenuse}$ therefore $\sin (θ) = \frac{\frac{\sqrt{39}}{8}}{1}$ .

$\frac{\frac{\sqrt{39}}{8}}{1}$

Step 4

Divide $\frac{\sqrt{39}}{8}$ by $1$ .

$\sin (θ) = \frac{\sqrt{39}}{8}$

Step 5

Approximate the result.

$\sin (θ) = \frac{\sqrt{39}}{8} \approx 0.78062474$