Calculus Examples

$y = \sin (x)$ , $y = \cos (x)$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$\sin (x) = \cos (x)$

Step 1.2

Solve $\sin (x) = \cos (x)$ for $x$ .

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Step 1.2.1

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

$\frac{\sin (x)}{\cos (x)} = \frac{\cos (x)}{\cos (x)}$

Step 1.2.2

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

$\tan (x) = \frac{\cos (x)}{\cos (x)}$

Step 1.2.3

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

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Step 1.2.3.1

Cancel the common factor.

$\tan (x) = \frac{\cos (x)}{\cos (x)}$

Step 1.2.3.2

Rewrite the expression.

$\tan (x) = 1$

Step 1.2.4

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

$x = \arctan (1)$

Step 1.2.5

Simplify the right side.

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Step 1.2.5.1

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

$x = \frac{π}{4}$

Step 1.2.6

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 1.2.7

Simplify $π + \frac{π}{4}$ .

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Step 1.2.7.1

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 1.2.7.2

Combine fractions.

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Step 1.2.7.2.1

Combine $π$ and $\frac{4}{4}$ .

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

Step 1.2.7.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 1.2.7.3

Simplify the numerator.

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Step 1.2.7.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 1.2.7.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 1.2.8

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

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Step 1.2.8.1

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

$\frac{π}{| b |}$

Step 1.2.8.2

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

$\frac{π}{| 1 |}$

Step 1.2.8.3

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

$\frac{π}{1}$

Step 1.2.8.4

Divide $π$ by $1$ .

$π$

Step 1.2.9

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 1.3

Evaluate $y$ when $x = \frac{π}{4} + π n$ .

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Step 1.3.1

Substitute $\frac{π}{4} + π n$ for $x$ .

$y = \cos (\frac{π}{4} + π n)$

Step 1.3.2

Remove parentheses.

$y = \cos (\frac{π}{4} + π n)$

Step 1.4

Evaluate $y$ when $x = \frac{5 π}{4} + π n$ .

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Step 1.4.1

Substitute $\frac{5 π}{4} + π n$ for $x$ .

$y = \cos (\frac{5 π}{4} + π n)$

Step 1.4.2

Remove parentheses.

$y = \cos (\frac{5 π}{4} + π n)$

Step 1.5

List all of the solutions.

$y = \cos (\frac{π}{4} + π n), x = \frac{π}{4} + π n$

$y = \cos (\frac{5 π}{4} + π n), x = \frac{5 π}{4} + π n$

$y = \cos (\frac{π}{4} + π n), x = \frac{π}{4} + π n$

$y = \cos (\frac{5 π}{4} + π n), x = \frac{5 π}{4} + π n$

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3