Trigonometry Examples

$\sin (v) = \sin (\frac{15 π}{14})$

Step 1

For the two functions to be equal, the arguments of each must be equal.

$v = \frac{15 π}{14}$

Step 2

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$v = π - \frac{15 π}{14}$

Step 3

Simplify the expression to find the second solution.

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Step 3.1

Add $2 π$ to $π - \frac{15 π}{14}$ .

$v = π - \frac{15 π}{14} + 2 π$

Step 3.2

The resulting angle of $\frac{27 π}{14}$ is positive and coterminal with $π - \frac{15 π}{14}$ .

$v = \frac{27 π}{14}$

Step 4

Find the period of $\sin (v)$ .

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Step 4.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.4

Divide $2 π$ by $1$ .

$2 π$

Step 5

The period of the $\sin (v)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$v = \frac{15 π}{14} + 2 π n, \frac{27 π}{14} + 2 π n$ , for any integer $n$