Algebra Examples

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

Step 1

Subtract $\sin^{2} (θ)$ from both sides of the equation.

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

Step 2

Simplify $(1 + \cos (θ)) (1 - \cos (θ)) - \sin^{2} (θ)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Expand $(1 + \cos (θ)) (1 - \cos (θ))$ using the FOIL Method.

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Step 2.1.1.1

Apply the distributive property.

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

Step 2.1.1.2

Apply the distributive property.

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

Step 2.1.1.3

Apply the distributive property.

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

Step 2.1.2

Simplify and combine like terms.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Multiply $1$ by $1$ .

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

Step 2.1.2.1.2

Multiply $- \cos (θ)$ by $1$ .

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

Step 2.1.2.1.3

Multiply $\cos (θ)$ by $1$ .

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

Step 2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 2.1.2.1.5

Multiply $- \cos (θ) \cos (θ)$ .

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Step 2.1.2.1.5.1

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

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

Step 2.1.2.1.5.2

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

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

Step 2.1.2.1.5.3

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

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

Step 2.1.2.1.5.4

Add $1$ and $1$ .

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

Step 2.1.2.2

Add $- \cos (θ)$ and $\cos (θ)$ .

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

Step 2.1.2.3

Add $1$ and $0$ .

$1 - \cos^{2} (θ) - \sin^{2} (θ) = 0$

Step 2.1.3

Apply pythagorean identity.

$\sin^{2} (θ) - \sin^{2} (θ) = 0$

Step 2.2

Subtract $\sin^{2} (θ)$ from $\sin^{2} (θ)$ .

$0 = 0$

Step 3

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

Always true

Step 4

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$