Trigonometry Examples

tan (θ) - 1 = 0

$\tan (θ) - 1 = 0$

Step 1

Add

1

$1$ to both sides of the equation.

tan (θ) = 1

$\tan (θ) = 1$

Step 2

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

θ

$θ$ from inside the tangent.

θ = arctan (1)

$θ = \arctan (1)$

Step 3

Simplify the right side.

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Step 3.1

The exact value of

arctan (1)

$\arctan (1)$ is

\frac{π}{4}

$\frac{π}{4}$ .

θ = \frac{π}{4}

$θ = \frac{π}{4}$

θ = \frac{π}{4}

$θ = \frac{π}{4}$

Step 4

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from

π

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

θ = π + \frac{π}{4}

$θ = π + \frac{π}{4}$

Step 5

Simplify

π + \frac{π}{4}

$π + \frac{π}{4}$ .

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Step 5.1

To write

π

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

\frac{4}{4}

$\frac{4}{4}$ .

θ = π \cdot \frac{4}{4} + \frac{π}{4}

$θ = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 5.2

Combine fractions.

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Step 5.2.1

Combine

π

$π$ and

\frac{4}{4}

$\frac{4}{4}$ .

θ = \frac{π \cdot 4}{4} + \frac{π}{4}

$θ = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 5.2.2

Combine the numerators over the common denominator.

θ = \frac{π \cdot 4 + π}{4}

$θ = \frac{π \cdot 4 + π}{4}$

θ = \frac{π \cdot 4 + π}{4}

$θ = \frac{π \cdot 4 + π}{4}$

Step 5.3

Simplify the numerator.

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Step 5.3.1

Move

4

$4$ to the left of

π

$π$ .

θ = \frac{4 \cdot π + π}{4}

$θ = \frac{4 \cdot π + π}{4}$

Step 5.3.2

Add

4 π

$4 π$ and

π

$π$ .

θ = \frac{5 π}{4}

$θ = \frac{5 π}{4}$

θ = \frac{5 π}{4}

$θ = \frac{5 π}{4}$

θ = \frac{5 π}{4}

$θ = \frac{5 π}{4}$

Step 6

Find the period of

tan (θ)

$\tan (θ)$ .

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Step 6.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 6.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

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{π}{1}

$\frac{π}{1}$

Step 6.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 7

The period of the

tan (θ)

$\tan (θ)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

θ = \frac{π}{4} + π n, \frac{5 π}{4} + π n

$θ = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer

$n$

Step 8

Consolidate the answers.

$θ = \frac{π}{4} + π n$ , for any integer

$n$