Trigonometry Examples

$1 - \sin (θ) = 2 \cos^{2} (θ)$

Step 1

Subtract $2 \cos^{2} (θ)$ from both sides of the equation.

$1 - \sin (θ) - 2 \cos^{2} (θ) = 0$

Step 2

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

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

Step 3

Simplify each term.

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Step 3.1

Apply the distributive property.

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

Step 3.2

Multiply $- 2$ by $1$ .

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

Step 3.3

Multiply $- 1$ by $- 2$ .

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

Step 4

Subtract $2$ from $1$ .

$- \sin (θ) - 1 + 2 \sin^{2} (θ) = 0$

Step 5

Reorder the polynomial.

$2 \sin^{2} (θ) - \sin (θ) - 1 = 0$

Step 6

Substitute $u$ for $\sin (θ)$ .

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

Step 7

Factor by grouping.

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Step 7.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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Step 7.1.1

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

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

Step 7.1.2

Rewrite $- 1$ as $1$ plus $- 2$

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

Step 7.1.3

Apply the distributive property.

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

Step 7.1.4

Multiply $u$ by $1$ .

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

Step 7.2

Factor out the greatest common factor from each group.

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Step 7.2.1

Group the first two terms and the last two terms.

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

Step 7.2.2

Factor out the greatest common factor (GCF) from each group.

$u (2 u + 1) - (2 u + 1) = 0$

Step 7.3

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$(2 u + 1) (u - 1) = 0$

Step 8

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

$2 u + 1 = 0$

$u - 1 = 0$

Step 9

Set $2 u + 1$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $2 u + 1$ equal to $0$ .

$2 u + 1 = 0$

Step 9.2

Solve $2 u + 1 = 0$ for $u$ .

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Step 9.2.1

Subtract $1$ from both sides of the equation.

$2 u = - 1$

Step 9.2.2

Divide each term in $2 u = - 1$ by $2$ and simplify.

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Step 9.2.2.1

Divide each term in $2 u = - 1$ by $2$ .

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 9.2.2.2

Simplify the left side.

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Step 9.2.2.2.1

Cancel the common factor of $2$ .

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Step 9.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 9.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{- 1}{2}$

Step 9.2.2.3

Simplify the right side.

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Step 9.2.2.3.1

Move the negative in front of the fraction.

$u = - \frac{1}{2}$

Step 10

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 10.2

Add $1$ to both sides of the equation.

$u = 1$

Step 11

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

$u = - \frac{1}{2}, 1$

Step 12

Substitute $\sin (θ)$ for $u$ .

$\sin (θ) = - \frac{1}{2}, 1$

Step 13

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

$\sin (θ) = - \frac{1}{2}$

$\sin (θ) = 1$

Step 14

Solve for $θ$ in $\sin (θ) = - \frac{1}{2}$ .

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Step 14.1

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

$θ = \arcsin (- \frac{1}{2})$

Step 14.2

Simplify the right side.

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Step 14.2.1

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

$θ = - \frac{π}{6}$

Step 14.3

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 14.4

Simplify the expression to find the second solution.

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Step 14.4.1

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

$θ = 2 π + \frac{π}{6} + π - 2 π$

Step 14.4.2

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 14.5

Find the period of $\sin (θ)$ .

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Step 14.5.1

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

$\frac{2 π}{| b |}$

Step 14.5.2

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

$\frac{2 π}{| 1 |}$

Step 14.5.3

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

$\frac{2 π}{1}$

Step 14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 14.6

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

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Step 14.6.1

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

$- \frac{π}{6} + 2 π$

Step 14.6.2

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 14.6.3

Combine fractions.

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Step 14.6.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 14.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 14.6.4

Simplify the numerator.

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Step 14.6.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 14.6.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 14.6.5

List the new angles.

$θ = \frac{11 π}{6}$

Step 14.7

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 15

Solve for $θ$ in $\sin (θ) = 1$ .

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Step 15.1

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

$θ = \arcsin (1)$

Step 15.2

Simplify the right side.

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Step 15.2.1

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

$θ = \frac{π}{2}$

Step 15.3

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.

$θ = π - \frac{π}{2}$

Step 15.4

Simplify $π - \frac{π}{2}$ .

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Step 15.4.1

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

$θ = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 15.4.2

Combine fractions.

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Step 15.4.2.1

Combine $π$ and $\frac{2}{2}$ .

$θ = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 15.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Step 15.4.3

Simplify the numerator.

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Step 15.4.3.1

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Step 15.4.3.2

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

Step 15.5

Find the period of $\sin (θ)$ .

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Step 15.5.1

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

$\frac{2 π}{| b |}$

Step 15.5.2

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

$\frac{2 π}{| 1 |}$

Step 15.5.3

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

$\frac{2 π}{1}$

Step 15.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 15.6

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

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

Step 16

List all of the solutions.

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

Step 17

Consolidate the answers.

$θ = \frac{π}{2} + \frac{2 π n}{3}$ , for any integer $n$