Precalculus Examples

${(\frac{1}{5})}^{2} + \cos^{2} (t) = 1$

Step 1

Simplify each term.

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Step 1.1

Apply the product rule to $\frac{1}{5}$ .

$\frac{1^{2}}{5^{2}} + \cos^{2} (t) = 1$

Step 1.2

One to any power is one.

$\frac{1}{5^{2}} + \cos^{2} (t) = 1$

Step 1.3

Raise $5$ to the power of $2$ .

$\frac{1}{25} + \cos^{2} (t) = 1$

Step 2

Move all terms not containing $\cos (t)$ to the right side of the equation.

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Step 2.1

Subtract $\frac{1}{25}$ from both sides of the equation.

$\cos^{2} (t) = 1 - \frac{1}{25}$

Step 2.2

Write $1$ as a fraction with a common denominator.

$\cos^{2} (t) = \frac{25}{25} - \frac{1}{25}$

Step 2.3

Combine the numerators over the common denominator.

$\cos^{2} (t) = \frac{25 - 1}{25}$

Step 2.4

Subtract $1$ from $25$ .

$\cos^{2} (t) = \frac{24}{25}$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\cos (t) = \pm \sqrt{\frac{24}{25}}$

Step 4

Simplify $\pm \sqrt{\frac{24}{25}}$ .

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Step 4.1

Rewrite $\sqrt{\frac{24}{25}}$ as $\frac{\sqrt{24}}{\sqrt{25}}$ .

$\cos (t) = \pm \frac{\sqrt{24}}{\sqrt{25}}$

Step 4.2

Simplify the numerator.

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Step 4.2.1

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 4.2.1.1

Factor $4$ out of $24$ .

$\cos (t) = \pm \frac{\sqrt{4 (6)}}{\sqrt{25}}$

Step 4.2.1.2

Rewrite $4$ as $2^{2}$ .

$\cos (t) = \pm \frac{\sqrt{2^{2} \cdot 6}}{\sqrt{25}}$

Step 4.2.2

Pull terms out from under the radical.

$\cos (t) = \pm \frac{2 \sqrt{6}}{\sqrt{25}}$

Step 4.3

Simplify the denominator.

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Step 4.3.1

Rewrite $25$ as $5^{2}$ .

$\cos (t) = \pm \frac{2 \sqrt{6}}{\sqrt{5^{2}}}$

Step 4.3.2

Pull terms out from under the radical, assuming positive real numbers.

$\cos (t) = \pm \frac{2 \sqrt{6}}{5}$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$\cos (t) = \frac{2 \sqrt{6}}{5}$

Step 5.2

Next, use the negative value of the $\pm$ to find the second solution.

$\cos (t) = - \frac{2 \sqrt{6}}{5}$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

$\cos (t) = \frac{2 \sqrt{6}}{5}, - \frac{2 \sqrt{6}}{5}$

Step 6

Set up each of the solutions to solve for $t$ .

$\cos (t) = \frac{2 \sqrt{6}}{5}$

$\cos (t) = - \frac{2 \sqrt{6}}{5}$

Step 7

Solve for $t$ in $\cos (t) = \frac{2 \sqrt{6}}{5}$ .

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Step 7.1

Take the inverse cosine of both sides of the equation to extract $t$ from inside the cosine.

$t = \arccos (\frac{2 \sqrt{6}}{5})$

Step 7.2

Simplify the right side.

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Step 7.2.1

Evaluate $\arccos (\frac{2 \sqrt{6}}{5})$ .

$t = 0.20135792$

Step 7.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$t = 2 (3.14159265) - 0.20135792$

Step 7.4

Solve for $t$ .

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Step 7.4.1

Remove parentheses.

$t = 2 (3.14159265) - 0.20135792$

Step 7.4.2

Simplify $2 (3.14159265) - 0.20135792$ .

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Step 7.4.2.1

Multiply $2$ by $3.14159265$ .

$t = 6.2831853 - 0.20135792$

Step 7.4.2.2

Subtract $0.20135792$ from $6.2831853$ .

$t = 6.08182738$

Step 7.5

Find the period of $\cos (t)$ .

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Step 7.5.1

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

$\frac{2 π}{| b |}$

Step 7.5.2

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

$\frac{2 π}{| 1 |}$

Step 7.5.3

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

$\frac{2 π}{1}$

Step 7.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 7.6

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

$t = 0.20135792 + 2 π n, 6.08182738 + 2 π n$ , for any integer $n$

Step 8

Solve for $t$ in $\cos (t) = - \frac{2 \sqrt{6}}{5}$ .

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Step 8.1

Take the inverse cosine of both sides of the equation to extract $t$ from inside the cosine.

$t = \arccos (- \frac{2 \sqrt{6}}{5})$

Step 8.2

Simplify the right side.

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Step 8.2.1

Evaluate $\arccos (- \frac{2 \sqrt{6}}{5})$ .

$t = 2.94023473$

Step 8.3

The cosine 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.

$t = 2 (3.14159265) - 2.94023473$

Step 8.4

Solve for $t$ .

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Step 8.4.1

Remove parentheses.

$t = 2 (3.14159265) - 2.94023473$

Step 8.4.2

Simplify $2 (3.14159265) - 2.94023473$ .

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Step 8.4.2.1

Multiply $2$ by $3.14159265$ .

$t = 6.2831853 - 2.94023473$

Step 8.4.2.2

Subtract $2.94023473$ from $6.2831853$ .

$t = 3.34295057$

Step 8.5

Find the period of $\cos (t)$ .

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Step 8.5.1

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

$\frac{2 π}{| b |}$

Step 8.5.2

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

$\frac{2 π}{| 1 |}$

Step 8.5.3

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

$\frac{2 π}{1}$

Step 8.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 8.6

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

$t = 2.94023473 + 2 π n, 3.34295057 + 2 π n$ , for any integer $n$

Step 9

List all of the solutions.

$t = 0.20135792 + 2 π n, 6.08182738 + 2 π n, 2.94023473 + 2 π n, 3.34295057 + 2 π n$ , for any integer $n$

Step 10

Consolidate the solutions.

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Step 10.1

Consolidate $0.20135792 + 2 π n$ and $3.34295057 + 2 π n$ to $0.20135792 + π n$ .

$t = 0.20135792 + π n, 6.08182738 + 2 π n, 2.94023473 + 2 π n$ , for any integer $n$

Step 10.2

Consolidate $6.08182738 + 2 π n$ and $2.94023473 + 2 π n$ to $2.94023473 + π n$ .

$t = 0.20135792 + π n, 2.94023473 + π n$ , for any integer $n$