Precalculus Examples

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

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $\sin (θ)$ from both sides of the equation.

$\cos^{2} (θ) - \sin^{2} (θ) - \sin (θ) = 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$\cos^{2} (θ) - \sin^{2} (θ) - \sin (θ) - 1 = 0$

Step 2

Simplify $\cos^{2} (θ) - \sin^{2} (θ) - \sin (θ) - 1$ .

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Step 2.1

Move $- 1$ .

$\cos^{2} (θ) - 1 - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.2

Reorder $\cos^{2} (θ)$ and $- 1$ .

$- 1 + \cos^{2} (θ) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.3

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

$- 1 (1) + \cos^{2} (θ) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.4

Factor $- 1$ out of $\cos^{2} (θ)$ .

$- 1 (1) - 1 (- \cos^{2} (θ)) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.5

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (θ))$ .

$- 1 (1 - \cos^{2} (θ)) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.6

Rewrite $- 1 (1 - \cos^{2} (θ))$ as $- (1 - \cos^{2} (θ))$ .

$- (1 - \cos^{2} (θ)) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.7

Apply pythagorean identity.

$- \sin^{2} (θ) - \sin^{2} (θ) - \sin (θ) = 0$

Step 2.8

Subtract $\sin^{2} (θ)$ from $- \sin^{2} (θ)$ .

$- 2 \sin^{2} (θ) - \sin (θ) = 0$

Step 3

Factor $- \sin (θ)$ out of $- 2 \sin^{2} (θ) - \sin (θ)$ .

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Step 3.1

Factor $- \sin (θ)$ out of $- 2 \sin^{2} (θ)$ .

$- \sin (θ) (2 \sin (θ)) - \sin (θ) = 0$

Step 3.2

Factor $- \sin (θ)$ out of $- \sin (θ)$ .

$- \sin (θ) (2 \sin (θ)) - \sin (θ) \cdot 1 = 0$

Step 3.3

Factor $- \sin (θ)$ out of $- \sin (θ) (2 \sin (θ)) - \sin (θ) \cdot 1$ .

$- \sin (θ) (2 \sin (θ) + 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (θ) = 0$

$2 \sin (θ) + 1 = 0$

Step 5

Set $\sin (θ)$ equal to $0$ and solve for $θ$ .

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Step 5.1

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Step 5.2

Solve $\sin (θ) = 0$ for $θ$ .

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Step 5.2.1

Take the inverse sine of both sides of the equation to extract $θ$ from inside the sine.

$θ = \arcsin (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$θ = 0$

Step 5.2.3

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

$θ = π - 0$

Step 5.2.4

Subtract $0$ from $π$ .

$θ = π$

Step 5.2.5

Find the period of $\sin (θ)$ .

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Step 5.2.5.1

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

$\frac{2 π}{| b |}$

Step 5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

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

$\frac{2 π}{1}$

Step 5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.6

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

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

Set $2 \sin (θ) + 1$ equal to $0$ and solve for $θ$ .

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Step 6.1

Set $2 \sin (θ) + 1$ equal to $0$ .

$2 \sin (θ) + 1 = 0$

Step 6.2

Solve $2 \sin (θ) + 1 = 0$ for $θ$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$2 \sin (θ) = - 1$

Step 6.2.2

Divide each term in $2 \sin (θ) = - 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 \sin (θ) = - 1$ by $2$ .

$\frac{2 \sin (θ)}{2} = \frac{- 1}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{- 1}{2}$

Step 6.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- 1}{2}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

$\sin (θ) = - \frac{1}{2}$

Step 6.2.3

Take the inverse sine of both sides of the equation to extract $θ$ from inside the sine.

$θ = \arcsin (- \frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$θ = - \frac{π}{6}$

Step 6.2.5

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{6} + π$

Step 6.2.6

Simplify the expression to find the second solution.

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Step 6.2.6.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$θ = 2 π + \frac{π}{6} + π - 2 π$

Step 6.2.6.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$θ = \frac{7 π}{6}$

Step 6.2.7

Find the period of $\sin (θ)$ .

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Step 6.2.7.1

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

$\frac{2 π}{| b |}$

Step 6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.7.3

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

$\frac{2 π}{1}$

Step 6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8

Add $2 π$ to every negative angle to get positive angles.

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Step 6.2.8.1

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 6.2.8.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 6.2.8.3

Combine fractions.

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Step 6.2.8.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 6.2.8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 6.2.8.4

Simplify the numerator.

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Step 6.2.8.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 6.2.8.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 6.2.8.5

List the new angles.

$θ = \frac{11 π}{6}$

Step 6.2.9

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

$θ = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $- \sin (θ) (2 \sin (θ) + 1) = 0$ true.

$θ = 2 π n, π + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$