Trigonometry Examples

$5 \cos (t) = - 1$

Step 1

Divide each term in $5 \cos (t) = - 1$ by $5$ and simplify.

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Divide each term in $5 \cos (t) = - 1$ by $5$ .

$\frac{5 \cos (t)}{5} = \frac{- 1}{5}$

Simplify the left side.

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Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 \cos (t)}{5} = \frac{- 1}{5}$

Divide $\cos (t)$ by $1$ .

$\cos (t) = \frac{- 1}{5}$

Simplify the right side.

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Move the negative in front of the fraction.

$\cos (t) = - \frac{1}{5}$

Step 2

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

$t = \arccos (- \frac{1}{5})$

Step 3

Simplify the right side.

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Evaluate $\arccos (- \frac{1}{5})$ .

$t = 1.77215424$

Step 4

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.

$t = 2 (3.14159265) - 1.77215424$

Step 5

Simplify $2 (3.14159265) - 1.77215424$ .

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Multiply $2$ by $3.14159265$ .

$t = 6.2831853 - 1.77215424$

Subtract $1.77215424$ from $6.2831853$ .

$t = 4.51103105$

Step 6

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

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Step 7

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

$t = 1.77215424 + 2 π n, 4.51103105 + 2 π n$ , for any integer $n$