Trigonometry Examples

$\cos (π - a) \cdot \tan (\frac{3 π}{2} - a)$

Step 1

Set the argument in $\tan (\frac{3 π}{2} - a)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$\frac{3 π}{2} - a = \frac{π}{2} + π n$ , for any integer $n$

Step 2

Solve for $a$ .

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Step 2.1

Move all terms not containing $a$ to the right side of the equation.

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Step 2.1.1

Subtract $\frac{3 π}{2}$ from both sides of the equation.

$- a = \frac{π}{2} + π n - \frac{3 π}{2}$

Step 2.1.2

Combine the numerators over the common denominator.

$- a = π n + \frac{π - 3 π}{2}$

Step 2.1.3

Subtract $3 π$ from $π$ .

$- a = π n + \frac{- 2 π}{2}$

Step 2.1.4

Cancel the common factor of $- 2$ and $2$ .

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Step 2.1.4.1

Factor $2$ out of $- 2 π$ .

$- a = π n + \frac{2 (- π)}{2}$

Step 2.1.4.2

Cancel the common factors.

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Step 2.1.4.2.1

Factor $2$ out of $2$ .

$- a = π n + \frac{2 (- π)}{2 (1)}$

Step 2.1.4.2.2

Cancel the common factor.

$- a = π n + \frac{2 (- π)}{2 \cdot 1}$

Step 2.1.4.2.3

Rewrite the expression.

$- a = π n + \frac{- π}{1}$

Step 2.1.4.2.4

Divide $- π$ by $1$ .

$- a = π n - π$

Step 2.2

Divide each term in $- a = π n - π$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- a = π n - π$ by $- 1$ .

$\frac{- a}{- 1} = \frac{π n}{- 1} + \frac{- π}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{a}{1} = \frac{π n}{- 1} + \frac{- π}{- 1}$

Step 2.2.2.2

Divide $a$ by $1$ .

$a = \frac{π n}{- 1} + \frac{- π}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Simplify each term.

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Step 2.2.3.1.1

Move the negative one from the denominator of $\frac{π n}{- 1}$ .

$a = - 1 \cdot (π n) + \frac{- π}{- 1}$

Step 2.2.3.1.2

Rewrite $- 1 \cdot (π n)$ as $- (π n)$ .

$a = - (π n) + \frac{- π}{- 1}$

Step 2.2.3.1.3

Dividing two negative values results in a positive value.

$a = - π n + \frac{π}{1}$

Step 2.2.3.1.4

Divide $π$ by $1$ .

$a = - π n + π$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

${a | a = - π n + π}$ $n$ , for any integer $n$

Step 4