Trigonometry Examples

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

Step 1

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

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

Step 2

Simplify each term.

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Step 2.1

Rewrite in terms of sines and cosines, then cancel the common factors.

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Step 2.1.1

Reorder $\sin (θ)$ and $\cot (θ)$ .

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

Step 2.1.2

Rewrite $\sin (θ) \cot (θ)$ in terms of sines and cosines.

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

Step 2.1.3

Cancel the common factors.

$\cos (θ) - (1 - \sin^{2} (θ)) = 0$

Step 2.2

Apply pythagorean identity.

$\cos (θ) - \cos^{2} (θ) = 0$

Step 3

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

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Step 3.1

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

$\cos (θ) \cdot 1 - \cos^{2} (θ) = 0$

Step 3.2

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

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

Step 3.3

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

$\cos (θ) (1 - \cos (θ)) = 0$

Step 4

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

$\cos (θ) = 0$

$1 - \cos (θ) = 0$

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 $1 - \cos (θ)$ equal to $0$ and solve for $θ$ .

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Step 6.1

Set $1 - \cos (θ)$ equal to $0$ .

$1 - \cos (θ) = 0$

Step 6.2

Solve $1 - \cos (θ) = 0$ for $θ$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$- \cos (θ) = - 1$

Step 6.2.2

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

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Step 6.2.2.1

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

$\frac{- \cos (θ)}{- 1} = \frac{- 1}{- 1}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\cos (θ)}{1} = \frac{- 1}{- 1}$

Step 6.2.2.2.2

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{- 1}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 1$ by $- 1$ .

$\cos (θ) = 1$

Step 6.2.3

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

$θ = \arccos (1)$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

Step 6.2.5

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.

$θ = 360 - 0$

Step 6.2.6

Subtract $0$ from $360$ .

$θ = 360$

Step 6.2.7

Find the period of $\cos (θ)$ .

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

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

Step 7

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

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

Step 8

Consolidate the answers.

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Step 8.1

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

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

Step 8.2

Consolidate $360 n$ and $360 + 360 n$ to $360 n$ .

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

Step 9

Exclude the solutions that do not make $\sin (θ) \cot (θ) - \cos^{2} (θ) = 0$ true.

$θ = 90, 270, 360, 450, 630, 720, 810, 990, 1080, 1170, 1350, 1440, 1800, 2160, 2520$ , for any integer $n$