Trigonometry Examples

$\cos (θ) - \tan (θ) \cos (θ) = 0$

Step 1

Simplify the left side.

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Step 1.1

Rewrite in terms of sines and cosines, then cancel the common factors.

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Step 1.1.1

Add parentheses.

$\cos (θ) - (\tan (θ) \cos (θ)) = 0$

Step 1.1.2

Rewrite $- \tan (θ) \cos (θ)$ in terms of sines and cosines.

$\cos (θ) - (\frac{\sin (θ)}{\cos (θ)} \cos (θ)) = 0$

Step 1.1.3

Cancel the common factors.

$\cos (θ) - \sin (θ) = 0$

Step 2

Divide each term in the equation by $\cos (θ)$ .

$\frac{\cos (θ)}{\cos (θ)} + \frac{- \sin (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Step 3

Cancel the common factor of $\cos (θ)$ .

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Step 3.1

Cancel the common factor.

$\frac{\cos (θ)}{\cos (θ)} + \frac{- \sin (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Step 3.2

Rewrite the expression.

$1 + \frac{- \sin (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Step 4

Separate fractions.

$1 + \frac{- 1}{1} \cdot \frac{\sin (θ)}{\cos (θ)} = \frac{0}{\cos (θ)}$

Step 5

Convert from $\frac{\sin (θ)}{\cos (θ)}$ to $\tan (θ)$ .

$1 + \frac{- 1}{1} \cdot \tan (θ) = \frac{0}{\cos (θ)}$

Step 6

Divide $- 1$ by $1$ .

$1 - \tan (θ) = \frac{0}{\cos (θ)}$

Step 7

Separate fractions.

$1 - \tan (θ) = \frac{0}{1} \cdot \frac{1}{\cos (θ)}$

Step 8

Convert from $\frac{1}{\cos (θ)}$ to $\sec (θ)$ .

$1 - \tan (θ) = \frac{0}{1} \cdot \sec (θ)$

Step 9

Divide $0$ by $1$ .

$1 - \tan (θ) = 0 \sec (θ)$

Step 10

Multiply $0$ by $\sec (θ)$ .

$1 - \tan (θ) = 0$

Step 11

Subtract $1$ from both sides of the equation.

$- \tan (θ) = - 1$

Step 12

Divide each term in $- \tan (θ) = - 1$ by $- 1$ and simplify.

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Step 12.1

Divide each term in $- \tan (θ) = - 1$ by $- 1$ .

$\frac{- \tan (θ)}{- 1} = \frac{- 1}{- 1}$

Step 12.2

Simplify the left side.

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Step 12.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (θ)}{1} = \frac{- 1}{- 1}$

Step 12.2.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{- 1}{- 1}$

Step 12.3

Simplify the right side.

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Step 12.3.1

Divide $- 1$ by $- 1$ .

$\tan (θ) = 1$

Step 13

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (1)$

Step 14

Simplify the right side.

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Step 14.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

Step 15

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

$θ = π + \frac{π}{4}$

Step 16

Simplify $π + \frac{π}{4}$ .

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Step 16.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 16.2

Combine fractions.

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Step 16.2.1

Combine $π$ and $\frac{4}{4}$ .

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

Step 16.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 + π}{4}$

Step 16.3

Simplify the numerator.

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Step 16.3.1

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π + π}{4}$

Step 16.3.2

Add $4 π$ and $π$ .

$θ = \frac{5 π}{4}$

Step 17

Find the period of $\tan (θ)$ .

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Step 17.1

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

$\frac{π}{| b |}$

Step 17.2

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

$\frac{π}{| 1 |}$

Step 17.3

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

$\frac{π}{1}$

Step 17.4

Divide $π$ by $1$ .

$π$

Step 18

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

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

Step 19

Consolidate the answers.

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