Trigonometry Examples

tan (θ) = sin (θ)

$\tan (θ) = \sin (θ)$

Step 1

Divide each term in

tan (θ) = sin (θ)

$\tan (θ) = \sin (θ)$ by

tan (θ)

$\tan (θ)$ and simplify.

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Step 1.1

Divide each term in

tan (θ) = sin (θ)

$\tan (θ) = \sin (θ)$ by

tan (θ)

$\tan (θ)$ .

\frac{tan (θ)}{tan (θ)} = \frac{sin (θ)}{tan (θ)}

$\frac{\tan (θ)}{\tan (θ)} = \frac{\sin (θ)}{\tan (θ)}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of

tan (θ)

$\tan (θ)$ .

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Step 1.2.1.1

Cancel the common factor.

\frac{tan (θ)}{tan (θ)} = \frac{sin (θ)}{tan (θ)}

$\frac{\tan (θ)}{\tan (θ)} = \frac{\sin (θ)}{\tan (θ)}$

Step 1.2.1.2

Rewrite the expression.

1 = \frac{sin (θ)}{tan (θ)}

$1 = \frac{\sin (θ)}{\tan (θ)}$

1 = \frac{sin (θ)}{tan (θ)}

$1 = \frac{\sin (θ)}{\tan (θ)}$

1 = \frac{sin (θ)}{tan (θ)}

$1 = \frac{\sin (θ)}{\tan (θ)}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Rewrite

tan (θ)

$\tan (θ)$ in terms of sines and cosines.

1 = \frac{sin (θ)}{\frac{sin (θ)}{cos (θ)}}

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

Step 1.3.2

Multiply by the reciprocal of the fraction to divide by

\frac{sin (θ)}{cos (θ)}

$\frac{\sin (θ)}{\cos (θ)}$ .

1 = sin (θ) \frac{cos (θ)}{sin (θ)}

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

Step 1.3.3

Write

sin (θ)

$\sin (θ)$ as a fraction with denominator

1

$1$ .

1 = \frac{sin (θ)}{1} \cdot \frac{cos (θ)}{sin (θ)}

$1 = \frac{\sin (θ)}{1} \cdot \frac{\cos (θ)}{\sin (θ)}$

Step 1.3.4

Cancel the common factor of

sin (θ)

$\sin (θ)$ .

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Step 1.3.4.1

Cancel the common factor.

1 = \frac{sin (θ)}{1} \cdot \frac{cos (θ)}{sin (θ)}

$1 = \frac{\sin (θ)}{1} \cdot \frac{\cos (θ)}{\sin (θ)}$

Step 1.3.4.2

Rewrite the expression.

1 = cos (θ)

$1 = \cos (θ)$

1 = cos (θ)

$1 = \cos (θ)$

1 = cos (θ)

$1 = \cos (θ)$

1 = cos (θ)

$1 = \cos (θ)$

Step 2

Rewrite the equation as

cos (θ) = 1

$\cos (θ) = 1$ .

cos (θ) = 1

$\cos (θ) = 1$

Step 3

Take the inverse cosine of both sides of the equation to extract

θ

$θ$ from inside the cosine.

θ = arccos (1)

$θ = \arccos (1)$

Step 4

Simplify the right side.

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Step 4.1

The exact value of

arccos (1)

$\arccos (1)$ is

0

$0$ .

θ = 0

$θ = 0$

θ = 0

$θ = 0$

Step 5

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

2 π

$2 π$ to find the solution in the fourth quadrant.

θ = 2 π - 0

$θ = 2 π - 0$

Step 6

Subtract

0

$0$ from

2 π

$2 π$ .

θ = 2 π

$θ = 2 π$

Step 7

Find the period of

cos (θ)

$\cos (θ)$ .

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Step 7.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 8

The period of the

cos (θ)

$\cos (θ)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

θ = 2 π n, 2 π + 2 π n

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

n

$n$

Step 9

Consolidate the answers.

θ = 2 π n

$θ = 2 π n$ , for any integer

n

$n$