Trigonometry Examples

$\sec (x) - \cos (x) = \tan (x) \sin (x)$

Step 1

Simplify the left side.

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Rewrite $\sec (x)$ in terms of sines and cosines.

$\frac{1}{\cos (x)} - \cos (x) = \tan (x) \sin (x)$

Step 2

Simplify the right side.

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Simplify $\tan (x) \sin (x)$ .

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

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

Multiply $\frac{\sin (x)}{\cos (x)} \sin (x)$ .

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Combine $\frac{\sin (x)}{\cos (x)}$ and $\sin (x)$ .

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

Raise $\sin (x)$ to the power of $1$ .

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

Raise $\sin (x)$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Step 3

Multiply both sides of the equation by $\cos (x)$ .

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

Step 4

Apply the distributive property.

$\cos (x) \frac{1}{\cos (x)} + \cos (x) (- \cos (x)) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 5

Cancel the common factor of $\cos (x)$ .

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Cancel the common factor.

$\cos (x) \frac{1}{\cos (x)} + \cos (x) (- \cos (x)) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Rewrite the expression.

$1 + \cos (x) (- \cos (x)) = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Step 6

Rewrite using the commutative property of multiplication.

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

Step 7

Multiply $- \cos (x) \cos (x)$ .

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Raise $\cos (x)$ to the power of $1$ .

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

Raise $\cos (x)$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1 - {\cos (x)}^{1 + 1} = \cos (x) \frac{\sin^{2} (x)}{\cos (x)}$

Add $1$ and $1$ .

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

Step 8

Apply pythagorean identity.

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

Step 9

Cancel the common factor of $\cos (x)$ .

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Cancel the common factor.

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

Rewrite the expression.

$\sin^{2} (x) = \sin^{2} (x)$

Step 10

Since the exponents are equal, the bases of the exponents on both sides of the equation must be equal.

$| \sin (x) | = | \sin (x) |$

Step 11

Solve for $x$ .

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Rewrite the absolute value equation as four equations without absolute value bars.

$\sin (x) = \sin (x)$

$\sin (x) = - \sin (x)$

$- \sin (x) = \sin (x)$

$- \sin (x) = - \sin (x)$

After simplifying, there are only two unique equations to be solved.

$\sin (x) = \sin (x)$

$\sin (x) = - \sin (x)$

Solve $\sin (x) = \sin (x)$ for $x$ .

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For the two functions to be equal, the arguments of each must be equal.

$x = x$

Move all terms containing $x$ to the left side of the equation.

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Subtract $x$ from both sides of the equation.

$x - x = 0$

Subtract $x$ from $x$ .

$0 = 0$

Since $0 = 0$ , the equation will always be true.

All real numbers

Solve $\sin (x) = - \sin (x)$ for $x$ .

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Move all terms containing $\sin (x)$ to the left side of the equation.

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

$\sin (x) + \sin (x) = 0$

Add $\sin (x)$ and $\sin (x)$ .

$2 \sin (x) = 0$

Divide each term in $2 \sin (x) = 0$ by $2$ and simplify.

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Divide each term in $2 \sin (x) = 0$ by $2$ .

$\frac{2 \sin (x)}{2} = \frac{0}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \sin (x)}{2} = \frac{0}{2}$

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{0}{2}$

Simplify the right side.

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Divide $0$ by $2$ .

$\sin (x) = 0$

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

$x = \arcsin (0)$

Simplify the right side.

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The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

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.

$x = π - 0$

Subtract $0$ from $π$ .

$x = π$

Find the period of $\sin (x)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Step 12

Consolidate the answers.

$x = π n$ , for any integer $n$