Trigonometry Examples

$(\frac{40}{41}, - \frac{9}{41})$

Step 1

To find the $\sin (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\frac{40}{41}, - \frac{9}{41})$ , draw the triangle between the three points $(0, 0)$ , $(\frac{40}{41}, 0)$ , and $(\frac{40}{41}, - \frac{9}{41})$ .

Opposite : $- \frac{9}{41}$

Adjacent : $\frac{40}{41}$

Step 2

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

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Apply the product rule to $\frac{40}{41}$ .

$\sqrt{\frac{40^{2}}{41^{2}} + {(- \frac{9}{41})}^{2}}$

Raise $40$ to the power of $2$ .

$\sqrt{\frac{1600}{41^{2}} + {(- \frac{9}{41})}^{2}}$

Raise $41$ to the power of $2$ .

$\sqrt{\frac{1600}{1681} + {(- \frac{9}{41})}^{2}}$

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

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Apply the product rule to $- \frac{9}{41}$ .

$\sqrt{\frac{1600}{1681} + {(- 1)}^{2} {(\frac{9}{41})}^{2}}$

Apply the product rule to $\frac{9}{41}$ .

$\sqrt{\frac{1600}{1681} + {(- 1)}^{2} \frac{9^{2}}{41^{2}}}$

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

$\sqrt{\frac{1600}{1681} + 1 \frac{9^{2}}{41^{2}}}$

Multiply $\frac{9^{2}}{41^{2}}$ by $1$ .

$\sqrt{\frac{1600}{1681} + \frac{9^{2}}{41^{2}}}$

Raise $9$ to the power of $2$ .

$\sqrt{\frac{1600}{1681} + \frac{81}{41^{2}}}$

Raise $41$ to the power of $2$ .

$\sqrt{\frac{1600}{1681} + \frac{81}{1681}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{1600 + 81}{1681}}$

Add $1600$ and $81$ .

$\sqrt{\frac{1681}{1681}}$

Divide $1681$ by $1681$ .

$\sqrt{1}$

Any root of $1$ is $1$ .

$1$

Step 3

$\sin (θ) = \frac{Opposite}{Hypotenuse}$ therefore $\sin (θ) = \frac{- \frac{9}{41}}{1}$ .

$\frac{- \frac{9}{41}}{1}$

Step 4

Divide $- \frac{9}{41}$ by $1$ .

$\sin (θ) = - \frac{9}{41}$

Step 5

Approximate the result.

$\sin (θ) = - \frac{9}{41} \approx - 0.\overline{21951}$