Trigonometry Examples

$\sin (2 θ) = \cos (θ)$

Step 1

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

$\sin (2 θ) - \cos (θ) = 0$

Step 2

Apply the sine double-angle identity.

$2 \sin (θ) \cos (θ) - \cos (θ) = 0$

Step 3

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

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Step 3.1

Factor $\cos (θ)$ out of $2 \sin (θ) \cos (θ)$ .

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

Step 3.2

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

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

Step 3.3

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

$\cos (θ) (2 \sin (θ) - 1) = 0$

Step 4

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

$\cos (θ) = 0$

$2 \sin (θ) - 1 = 0$

Step 5

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

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Step 5.1

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Step 5.2

Solve $\cos (θ) = 0$ for $θ$ .

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Step 5.2.1

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

$θ = \arccos (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arccos (0)$ is $90$ .

$θ = 90$

Step 5.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 90$

Step 5.2.4

Subtract $90$ from $360$ .

$θ = 270$

Step 5.2.5

Find the period of $\cos (θ)$ .

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Step 5.2.5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 5.2.5.2

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

$\frac{360}{| 1 |}$

Step 5.2.5.3

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

$\frac{360}{1}$

Step 5.2.5.4

Divide $360$ by $1$ .

$360$

Step 5.2.6

The period of the $\cos (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 90 + 360 n, 270 + 360 n$ , 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

Add $1$ to 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 6.2.3

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 $30$ .

$θ = 30$

Step 6.2.5

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$θ = 180 - 30$

Step 6.2.6

Subtract $30$ from $180$ .

$θ = 150$

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 $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 6.2.7.2

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

$\frac{360}{| 1 |}$

Step 6.2.7.3

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

$\frac{360}{1}$

Step 6.2.7.4

Divide $360$ by $1$ .

$360$

Step 6.2.8

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 30 + 360 n, 150 + 360 n$ , for any integer $n$

Step 7

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

$θ = 90 + 360 n, 270 + 360 n, 30 + 360 n, 150 + 360 n$ , for any integer $n$

Step 8

Consolidate $90 + 360 n$ and $270 + 360 n$ to $90 + 180 n$ .

$θ = 90 + 180 n, 30 + 360 n, 150 + 360 n$ , for any integer $n$