Trigonometry Examples

$\sin^{2} (x) = 2 \cos (x) + 2$

Step 1

Replace the $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$1 - \cos^{2} (x) = 2 \cos (x) + 2$

Step 2

Reorder the polynomial.

$- \cos^{2} (x) + 1 = 2 \cos (x) + 2$

Step 3

Substitute $u$ for $\cos (x)$ .

$- {(u)}^{2} + 1 = 2 (u) + 2$

Step 4

Subtract $2 u$ from both sides of the equation.

$- u^{2} + 1 - 2 u = 2$

Step 5

Subtract $2$ from both sides of the equation.

$- u^{2} + 1 - 2 u - 2 = 0$

Step 6

Subtract $2$ from $1$ .

$- u^{2} - 2 u - 1 = 0$

Step 7

Factor the left side of the equation.

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Step 7.1

Factor $- 1$ out of $- u^{2} - 2 u - 1$ .

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Step 7.1.1

Factor $- 1$ out of $- u^{2}$ .

$- u^{2} - 2 u - 1 = 0$

Step 7.1.2

Factor $- 1$ out of $- 2 u$ .

$- u^{2} - (2 u) - 1 = 0$

Step 7.1.3

Rewrite $- 1$ as $- 1 (1)$ .

$- u^{2} - (2 u) - 1 \cdot 1 = 0$

Step 7.1.4

Factor $- 1$ out of $- u^{2} - (2 u)$ .

$- (u^{2} + 2 u) - 1 \cdot 1 = 0$

Step 7.1.5

Factor $- 1$ out of $- (u^{2} + 2 u) - 1 (1)$ .

$- (u^{2} + 2 u + 1) = 0$

Step 7.2

Factor using the perfect square rule.

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Step 7.2.1

Rewrite $1$ as $1^{2}$ .

$- (u^{2} + 2 u + 1^{2}) = 0$

Step 7.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 7.2.3

Rewrite the polynomial.

$- (u^{2} + 2 \cdot u \cdot 1 + 1^{2}) = 0$

Step 7.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 1$ .

$- {(u + 1)}^{2} = 0$

Step 8

Divide each term in $- {(u + 1)}^{2} = 0$ by $- 1$ and simplify.

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Step 8.1

Divide each term in $- {(u + 1)}^{2} = 0$ by $- 1$ .

$\frac{- {(u + 1)}^{2}}{- 1} = \frac{0}{- 1}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Dividing two negative values results in a positive value.

$\frac{{(u + 1)}^{2}}{1} = \frac{0}{- 1}$

Step 8.2.2

Divide ${(u + 1)}^{2}$ by $1$ .

${(u + 1)}^{2} = \frac{0}{- 1}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Divide $0$ by $- 1$ .

${(u + 1)}^{2} = 0$

Step 9

Set the $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 10

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 11

Substitute $\cos (x)$ for $u$ .

$\cos (x) = - 1$

Step 12

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

$x = \arccos (- 1)$

Step 13

Simplify the right side.

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Step 13.1

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

$x = π$

Step 14

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.

$x = 2 π - π$

Step 15

Subtract $π$ from $2 π$ .

$x = π$

Step 16

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

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Step 16.1

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

$\frac{2 π}{| b |}$

Step 16.2

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

$\frac{2 π}{| 1 |}$

Step 16.3

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

$\frac{2 π}{1}$

Step 16.4

Divide $2 π$ by $1$ .

$2 π$

Step 17

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

$x = π + 2 π n$ , for any integer $n$