Basic Math Examples

sin (\frac{π}{2} + θ) = - tan (θ)

$\sin (\frac{π}{2} + θ) = - \tan (θ)$

Step 1

Use the sum formula for sine to simplify the expression. The formula states that

sin (A + B) = sin (A) cos (B) + cos (A) sin (B)

$\sin (A + B) = \sin (A) \cos (B) + \cos (A) \sin (B)$ .

sin (\frac{π}{2}) cos (θ) + cos (\frac{π}{2}) sin (θ) = - tan (θ)

$\sin (\frac{π}{2}) \cos (θ) + \cos (\frac{π}{2}) \sin (θ) = - \tan (θ)$

Step 2

Simplify the left side.

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Step 2.1

Simplify

sin (\frac{π}{2}) cos (θ) + cos (\frac{π}{2}) sin (θ)

$\sin (\frac{π}{2}) \cos (θ) + \cos (\frac{π}{2}) \sin (θ)$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

The exact value of

sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

1 cos (θ) + cos (\frac{π}{2}) sin (θ) = - tan (θ)

$1 \cos (θ) + \cos (\frac{π}{2}) \sin (θ) = - \tan (θ)$

Step 2.1.1.2

Multiply

cos (θ)

$\cos (θ)$ by

1

$1$ .

cos (θ) + cos (\frac{π}{2}) sin (θ) = - tan (θ)

$\cos (θ) + \cos (\frac{π}{2}) \sin (θ) = - \tan (θ)$

Step 2.1.1.3

The exact value of

cos (\frac{π}{2})

$\cos (\frac{π}{2})$ is

0

$0$ .

cos (θ) + 0 sin (θ) = - tan (θ)

$\cos (θ) + 0 \sin (θ) = - \tan (θ)$

Step 2.1.1.4

Multiply

0

$0$ by

sin (θ)

$\sin (θ)$ .

cos (θ) + 0 = - tan (θ)

$\cos (θ) + 0 = - \tan (θ)$

cos (θ) + 0 = - tan (θ)

$\cos (θ) + 0 = - \tan (θ)$

Step 2.1.2

Add

cos (θ)

$\cos (θ)$ and

0

$0$ .

cos (θ) = - tan (θ)

$\cos (θ) = - \tan (θ)$

cos (θ) = - tan (θ)

$\cos (θ) = - \tan (θ)$

cos (θ) = - tan (θ)

$\cos (θ) = - \tan (θ)$

Step 3

Simplify the right side.

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Step 3.1

Rewrite

tan (θ)

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

cos (θ) = - \frac{sin (θ)}{cos (θ)}

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

cos (θ) = - \frac{sin (θ)}{cos (θ)}

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

Step 4

Multiply both sides of the equation by

cos (θ)

$\cos (θ)$ .

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

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

Step 5

Multiply

cos (θ) cos (θ)

$\cos (θ) \cos (θ)$ .

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Step 5.1

Raise

cos (θ)

$\cos (θ)$ to the power of

1

$1$ .

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

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

Step 5.2

Raise

cos (θ)

$\cos (θ)$ to the power of

1

$1$ .

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

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

Step 5.3

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

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

Step 5.4

Add

1

$1$ and

1

$1$ .

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

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

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

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

Step 6

Rewrite using the commutative property of multiplication.

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

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

Step 7

Cancel the common factor of

cos (θ)

$\cos (θ)$ .

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Step 7.1

Factor

cos (θ)

$\cos (θ)$ out of

- cos (θ)

$- \cos (θ)$ .

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

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

Step 7.2

Cancel the common factor.

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

Step 7.3

Rewrite the expression.

$\cos^{2} (θ) = - \sin (θ)$

Step 8

Add

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

$\cos^{2} (θ) + \sin (θ) = 0$

Step 9

Replace

$\cos^{2} (θ)$ with

$1 - \sin^{2} (θ)$ .

$(1 - \sin^{2} (θ)) + \sin (θ) = 0$

Step 10

Solve for

$θ$ .

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Step 10.1

Substitute

$u$ for

$\sin (θ)$ .

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

Step 10.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 10.3

Substitute the values

$a = - 1$ ,

$b = 1$ , and

$c = 1$ into the quadratic formula and solve for

$u$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (- 1 \cdot 1)}}{2 \cdot - 1}$

Step 10.4

Simplify.

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Step 10.4.1

Simplify the numerator.

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Step 10.4.1.1

One to any power is one.

$u = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 1}}{2 \cdot - 1}$

Step 10.4.1.2

Multiply

$- 4 \cdot - 1 \cdot 1$ .

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Step 10.4.1.2.1

Multiply

$- 4$ by

$- 1$ .

$u = \frac{- 1 \pm \sqrt{1 + 4 \cdot 1}}{2 \cdot - 1}$

Step 10.4.1.2.2

Multiply

$4$ by

$1$ .

$u = \frac{- 1 \pm \sqrt{1 + 4}}{2 \cdot - 1}$

Step 10.4.1.3

Add

$1$ and

$4$ .

$u = \frac{- 1 \pm \sqrt{5}}{2 \cdot - 1}$

Step 10.4.2

Multiply

$2$ by

$- 1$ .

$u = \frac{- 1 \pm \sqrt{5}}{- 2}$

Step 10.4.3

Simplify

$\frac{- 1 \pm \sqrt{5}}{- 2}$ .

$u = \frac{1 \pm \sqrt{5}}{2}$

Step 10.5

The final answer is the combination of both solutions.

$u = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 10.6

Substitute

$\sin (θ)$ for

$u$ .

$\sin (θ) = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 10.7

Set up each of the solutions to solve for

$θ$ .

$\sin (θ) = \frac{1 + \sqrt{5}}{2}$

$\sin (θ) = \frac{1 - \sqrt{5}}{2}$

Step 10.8

Solve for

$θ$ in

$\sin (θ) = \frac{1 + \sqrt{5}}{2}$ .

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Step 10.8.1

The range of sine is

$- 1 \leq y \leq 1$ . Since

$\frac{1 + \sqrt{5}}{2}$ does not fall in this range, there is no solution.

No solution

Step 10.9

Solve for

$θ$ in

$\sin (θ) = \frac{1 - \sqrt{5}}{2}$ .

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Step 10.9.1

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

$θ$ from inside the sine.

$θ = \arcsin (\frac{1 - \sqrt{5}}{2})$

Step 10.9.2

Simplify the right side.

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Step 10.9.2.1

Evaluate

$\arcsin (\frac{1 - \sqrt{5}}{2})$ .

$θ = - 0.66623943$

Step 10.9.3

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.

$θ = (3.14159265) + 0.66623943$

Step 10.9.4

Solve for

$θ$ .

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Step 10.9.4.1

Remove parentheses.

$θ = 3.14159265 + 0.66623943$

Step 10.9.4.2

Remove parentheses.

$θ = (3.14159265) + 0.66623943$

Step 10.9.4.3

Add

$3.14159265$ and

$0.66623943$ .

$θ = 3.80783208$

Step 10.9.5

Find the period of

$\sin (θ)$ .

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Step 10.9.5.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 10.9.5.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 10.9.5.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 10.9.5.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 10.9.6

Add

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

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Step 10.9.6.1

Add

$2 π$ to

$- 0.66623943$ to find the positive angle.

$- 0.66623943 + 2 π$

Step 10.9.6.2

Subtract

$0.66623943$ from

$2 π$ .

$5.61694587$

Step 10.9.6.3

List the new angles.

$θ = 5.61694587$

Step 10.9.7

The period of the

$\sin (θ)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$θ = 3.80783208 + 2 π n, 5.61694587 + 2 π n$ , for any integer

$n$

$θ = 3.80783208 + 2 π n, 5.61694587 + 2 π n$ , for any integer

$n$

Step 10.10

List all of the solutions.

$θ = 3.80783208 + 2 π n, 5.61694587 + 2 π n$ , for any integer

$n$

$θ = 3.80783208 + 2 π n, 5.61694587 + 2 π n$ , for any integer

$n$