Trigonometry Examples

2 \sin (θ) = \sqrt{3}

$2 \sin (θ) = \sqrt{3}$

Step 1

Divide each term in

2 \sin (θ) = \sqrt{3}

$2 \sin (θ) = \sqrt{3}$ by

2

$2$ and simplify.

Tap for more steps...

Step 1.1

Divide each term in

2 \sin (θ) = \sqrt{3}

$2 \sin (θ) = \sqrt{3}$ by

2

$2$ .

\frac{2 \sin (θ)}{2} = \frac{\sqrt{3}}{2}

$\frac{2 \sin (θ)}{2} = \frac{\sqrt{3}}{2}$

Step 1.2

Simplify the left side.

Tap for more steps...

Step 1.2.1

Cancel the common factor of

2

$2$ .

Tap for more steps...

Step 1.2.1.1

Cancel the common factor.

\frac{2 \sin (θ)}{2} = \frac{\sqrt{3}}{2}

$\frac{2 \sin (θ)}{2} = \frac{\sqrt{3}}{2}$

Step 1.2.1.2

Divide

\sin (θ)

$\sin (θ)$ by

1

$1$ .

\sin (θ) = \frac{\sqrt{3}}{2}

$\sin (θ) = \frac{\sqrt{3}}{2}$

\sin (θ) = \frac{\sqrt{3}}{2}

$\sin (θ) = \frac{\sqrt{3}}{2}$

\sin (θ) = \frac{\sqrt{3}}{2}

$\sin (θ) = \frac{\sqrt{3}}{2}$

\sin (θ) = \frac{\sqrt{3}}{2}

$\sin (θ) = \frac{\sqrt{3}}{2}$

Step 2

Take the inverse sine of both sides of the equation to extract

θ

$θ$ from inside the sine.

θ = \arcsin (\frac{\sqrt{3}}{2})

$θ = \arcsin (\frac{\sqrt{3}}{2})$

Step 3

Simplify the right side.

Tap for more steps...

Step 3.1

The exact value of

\arcsin (\frac{\sqrt{3}}{2})

$\arcsin (\frac{\sqrt{3}}{2})$ is

\frac{π}{3}

$\frac{π}{3}$ .

θ = \frac{π}{3}

$θ = \frac{π}{3}$

θ = \frac{π}{3}

$θ = \frac{π}{3}$

Step 4

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

π

$π$ to find the solution in the second quadrant.

θ = π - \frac{π}{3}

$θ = π - \frac{π}{3}$

Step 5

Simplify

π - \frac{π}{3}

$π - \frac{π}{3}$ .

Tap for more steps...

Step 5.1

To write

π

$π$ as a fraction with a common denominator, multiply by

\frac{3}{3}

$\frac{3}{3}$ .

θ = π \cdot \frac{3}{3} - \frac{π}{3}

$θ = π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.2

Combine fractions.

Tap for more steps...

Step 5.2.1

Combine

π

$π$ and

\frac{3}{3}

$\frac{3}{3}$ .

θ = \frac{π \cdot 3}{3} - \frac{π}{3}

$θ = \frac{π \cdot 3}{3} - \frac{π}{3}$

Step 5.2.2

Combine the numerators over the common denominator.

θ = \frac{π \cdot 3 - π}{3}

$θ = \frac{π \cdot 3 - π}{3}$

θ = \frac{π \cdot 3 - π}{3}

$θ = \frac{π \cdot 3 - π}{3}$

Step 5.3

Simplify the numerator.

Tap for more steps...

Step 5.3.1

Move

3

$3$ to the left of

π

$π$ .

θ = \frac{3 \cdot π - π}{3}

$θ = \frac{3 \cdot π - π}{3}$

Step 5.3.2

Subtract

π

$π$ from

3 π

$3 π$ .

θ = \frac{2 π}{3}

$θ = \frac{2 π}{3}$

θ = \frac{2 π}{3}

$θ = \frac{2 π}{3}$

θ = \frac{2 π}{3}

$θ = \frac{2 π}{3}$

Step 6

Find the period of

\sin (θ)

$\sin (θ)$ .

Tap for more steps...

Step 6.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 6.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 6.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 7

The period of the

\sin (θ)

$\sin (θ)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n$ , for any integer

n

$n$