Trigonometry Examples

$\cos (π - x) + \sin (\frac{π}{2} + 0) = 0$

Step 1

Divide each term in the equation by $\cos (π - x)$ .

$\frac{\cos (π - x)}{\cos (π - x)} + \frac{\sin (\frac{π}{2} + 0)}{\cos (π - x)} = \frac{0}{\cos (π - x)}$

Step 2

Cancel the common factor of $\cos (π - x)$ .

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Step 2.1

Cancel the common factor.

$\frac{\cos (π - x)}{\cos (π - x)} + \frac{\sin (\frac{π}{2} + 0)}{\cos (π - x)} = \frac{0}{\cos (π - x)}$

Step 2.2

Rewrite the expression.

$1 + \frac{\sin (\frac{π}{2} + 0)}{\cos (π - x)} = \frac{0}{\cos (π - x)}$

Step 3

Rewrite $\frac{\sin (\frac{π}{2} + 0)}{\cos (π - x)}$ as a product.

$1 + \sin (\frac{π}{2} + 0) (\frac{1}{\cos (π - x)}) = \frac{0}{\cos (π - x)}$

Step 4

Write $\sin (\frac{π}{2} + 0)$ as a fraction with denominator $1$ .

$1 + \frac{\sin (\frac{π}{2} + 0)}{1} \cdot \frac{1}{\cos (π - x)} = \frac{0}{\cos (π - x)}$

Step 5

Simplify.

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Step 5.1

Divide $\sin (\frac{π}{2} + 0)$ by $1$ .

$1 + \sin (\frac{π}{2} + 0) (\frac{1}{\cos (π - x)}) = \frac{0}{\cos (π - x)}$

Step 5.2

Convert from $\frac{1}{\cos (π - x)}$ to $\sec (π - x)$ .

$1 + \sin (\frac{π}{2} + 0) \sec (π - x) = \frac{0}{\cos (π - x)}$

Step 6

Add $\frac{π}{2}$ and $0$ .

$1 + \sin (\frac{π}{2}) \sec (π - x) = \frac{0}{\cos (π - x)}$

Step 7

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$1 + 1 \sec (π - x) = \frac{0}{\cos (π - x)}$

Step 8

Multiply $\sec (π - x)$ by $1$ .

$1 + \sec (π - x) = \frac{0}{\cos (π - x)}$

Step 9

Separate fractions.

$1 + \sec (π - x) = \frac{0}{1} \cdot \frac{1}{\cos (π - x)}$

Step 10

Convert from $\frac{1}{\cos (π - x)}$ to $\sec (π - x)$ .

$1 + \sec (π - x) = \frac{0}{1} \cdot \sec (π - x)$

Step 11

Divide $0$ by $1$ .

$1 + \sec (π - x) = 0 \sec (π - x)$

Step 12

Multiply $0$ by $\sec (π - x)$ .

$1 + \sec (π - x) = 0$

Step 13

Subtract $1$ from both sides of the equation.

$\sec (π - x) = - 1$

Step 14

Take the inverse secant of both sides of the equation to extract $x$ from inside the secant.

$π - x = arcsec (- 1)$

Step 15

Simplify the right side.

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Step 15.1

The exact value of $arcsec (- 1)$ is $π$ .

$π - x = π$

Step 16

Move all terms not containing $x$ to the right side of the equation.

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Step 16.1

Subtract $π$ from both sides of the equation.

$- x = π - π$

Step 16.2

Subtract $π$ from $π$ .

$- x = 0$

Step 17

Divide each term in $- x = 0$ by $- 1$ and simplify.

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Step 17.1

Divide each term in $- x = 0$ by $- 1$ .

$\frac{- x}{- 1} = \frac{0}{- 1}$

Step 17.2

Simplify the left side.

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Step 17.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{0}{- 1}$

Step 17.2.2

Divide $x$ by $1$ .

$x = \frac{0}{- 1}$

Step 17.3

Simplify the right side.

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Step 17.3.1

Divide $0$ by $- 1$ .

$x = 0$

Step 18

The secant function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$π - x = 2 π - π$

Step 19

Solve for $x$ .

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Step 19.1

Subtract $π$ from $2 π$ .

$π - x = π$

Step 19.2

Move all terms not containing $x$ to the right side of the equation.

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Step 19.2.1

Subtract $π$ from both sides of the equation.

$- x = π - π$

Step 19.2.2

Subtract $π$ from $π$ .

$- x = 0$

Step 19.3

Divide each term in $- x = 0$ by $- 1$ and simplify.

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Step 19.3.1

Divide each term in $- x = 0$ by $- 1$ .

$\frac{- x}{- 1} = \frac{0}{- 1}$

Step 19.3.2

Simplify the left side.

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Step 19.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{0}{- 1}$

Step 19.3.2.2

Divide $x$ by $1$ .

$x = \frac{0}{- 1}$

Step 19.3.3

Simplify the right side.

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Step 19.3.3.1

Divide $0$ by $- 1$ .

$x = 0$

Step 20

Find the period of $\sec (π - x)$ .

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Step 20.1

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

$\frac{2 π}{| b |}$

Step 20.2

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

$\frac{2 π}{| - 1 |}$

Step 20.3

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

$\frac{2 π}{1}$

Step 20.4

Divide $2 π$ by $1$ .

$2 π$

Step 21

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

$x = 2 π n$ , for any integer $n$