Trigonometry Examples

$\sin (x) - \cos (x) > 0$

Step 1

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

$\frac{\sin (x)}{\cos (x)} + \frac{- \cos (x)}{\cos (x)} > \frac{0}{\cos (x)}$

Step 2

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

$\tan (x) + \frac{- \cos (x)}{\cos (x)} > \frac{0}{\cos (x)}$

Step 3

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

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Step 3.1

Cancel the common factor.

$\tan (x) + \frac{- \cos (x)}{\cos (x)} > \frac{0}{\cos (x)}$

Step 3.2

Divide $- 1$ by $1$ .

$\tan (x) - 1 > \frac{0}{\cos (x)}$

Step 4

Separate fractions.

$\tan (x) - 1 > \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 5

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

$\tan (x) - 1 > \frac{0}{1} \cdot \sec (x)$

Step 6

Divide $0$ by $1$ .

$\tan (x) - 1 > 0 \sec (x)$

Step 7

Multiply $0$ by $\sec (x)$ .

$\tan (x) - 1 > 0$

Step 8

Add $1$ to both sides of the inequality.

$\tan (x) > 1$

Step 9

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

$x > \arctan (1)$

Step 10

Simplify the right side.

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Step 10.1

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

$x > \frac{π}{4}$

Step 11

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.

$x = π + \frac{π}{4}$

Step 12

Simplify $π + \frac{π}{4}$ .

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Step 12.1

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

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

Step 12.2

Combine fractions.

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Step 12.2.1

Combine $π$ and $\frac{4}{4}$ .

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

Step 12.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 12.3

Simplify the numerator.

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Step 12.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 12.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 13

Find the period of $\tan (x)$ .

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Step 13.1

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

$\frac{π}{| b |}$

Step 13.2

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

$\frac{π}{| 1 |}$

Step 13.3

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

$\frac{π}{1}$

Step 13.4

Divide $π$ by $1$ .

$π$

Step 14

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

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

Step 15

Consolidate the answers.

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

Step 16

Use each root to create test intervals.

$\frac{π}{4} < x < \frac{5 π}{4}$

Step 17

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 17.1

Test a value on the interval $\frac{π}{4} < x < \frac{5 π}{4}$ to see if it makes the inequality true.

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Step 17.1.1

Choose a value on the interval $\frac{π}{4} < x < \frac{5 π}{4}$ and see if this value makes the original inequality true.

$x = 2$

Step 17.1.2

Replace $x$ with $2$ in the original inequality.

$\sin (2) - \cos (2) > 0$

Step 17.1.3

The left side $1.32544426$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 17.2

Compare the intervals to determine which ones satisfy the original inequality.

$\frac{π}{4} < x < \frac{5 π}{4}$ True

Step 18

The solution consists of all of the true intervals.

$\frac{π}{4} + π n < x < \frac{5 π}{4} + π n$ , for any integer $n$

Step 19