Trigonometry Examples

\tan (θ) = 3

$\tan (θ) = 3$

Step 1

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

θ

$θ$ from inside the tangent.

θ = \arctan (3)

$θ = \arctan (3)$

Step 2

Simplify the right side.

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Step 2.1

Evaluate

\arctan (3)

$\arctan (3)$ .

θ = 1.24904577

$θ = 1.24904577$

θ = 1.24904577

$θ = 1.24904577$

Step 3

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 = (3.14159265) + 1.24904577

$θ = (3.14159265) + 1.24904577$

Step 4

Solve for

θ

$θ$ .

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Step 4.1

Remove parentheses.

θ = 3.14159265 + 1.24904577

$θ = 3.14159265 + 1.24904577$

Step 4.2

Remove parentheses.

θ = (3.14159265) + 1.24904577

$θ = (3.14159265) + 1.24904577$

Step 4.3

Add

3.14159265

$3.14159265$ and

1.24904577

$1.24904577$ .

θ = 4.39063842

$θ = 4.39063842$

θ = 4.39063842

$θ = 4.39063842$

Step 5

Find the period of

\tan (θ)

$\tan (θ)$ .

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Step 5.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{π}{| 1 |}

$\frac{π}{| 1 |}$

Step 5.3

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 5.4

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

Step 6

The period of the

\tan (θ)

$\tan (θ)$ function is

π

$π$ so values will repeat every

π

$π$ radians in both directions.

θ = 1.24904577 + π n, 4.39063842 + π n

$θ = 1.24904577 + π n, 4.39063842 + π n$ , for any integer

n

$n$

Step 7

Consolidate

1.24904577 + π n

$1.24904577 + π n$ and

4.39063842 + π n

$4.39063842 + π n$ to

1.24904577 + π n

$1.24904577 + π n$ .

θ = 1.24904577 + π n

$θ = 1.24904577 + π n$ , for any integer

n

$n$