Precalculus Examples

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

Step 1

Simplify the left side.

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Step 1.1

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

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Step 1.1.1

Simplify each term.

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Step 1.1.1.1

Rewrite ${(\sin (x) + \cos (x))}^{2}$ as $(\sin (x) + \cos (x)) (\sin (x) + \cos (x))$ .

$(\sin (x) + \cos (x)) (\sin (x) + \cos (x)) - 1 = \sin (2 x)$

Step 1.1.1.2

Expand $(\sin (x) + \cos (x)) (\sin (x) + \cos (x))$ using the FOIL Method.

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Step 1.1.1.2.1

Apply the distributive property.

$\sin (x) (\sin (x) + \cos (x)) + \cos (x) (\sin (x) + \cos (x)) - 1 = \sin (2 x)$

Step 1.1.1.2.2

Apply the distributive property.

$\sin (x) \sin (x) + \sin (x) \cos (x) + \cos (x) (\sin (x) + \cos (x)) - 1 = \sin (2 x)$

Step 1.1.1.2.3

Apply the distributive property.

$\sin (x) \sin (x) + \sin (x) \cos (x) + \cos (x) \sin (x) + \cos (x) \cos (x) - 1 = \sin (2 x)$

Step 1.1.1.3

Simplify and combine like terms.

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Step 1.1.1.3.1

Simplify each term.

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Step 1.1.1.3.1.1

Multiply $\sin (x) \sin (x)$ .

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Step 1.1.1.3.1.1.1

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

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

Step 1.1.1.3.1.1.2

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

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

Step 1.1.1.3.1.1.3

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

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

Step 1.1.1.3.1.1.4

Add $1$ and $1$ .

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

Step 1.1.1.3.1.2

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

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Step 1.1.1.3.1.2.1

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

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

Step 1.1.1.3.1.2.2

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

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

Step 1.1.1.3.1.2.3

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

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

Step 1.1.1.3.1.2.4

Add $1$ and $1$ .

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

Step 1.1.1.3.2

Reorder the factors of $\sin (x) \cos (x)$ .

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

Step 1.1.1.3.3

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

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

Step 1.1.1.4

Move $\cos^{2} (x)$ .

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

Step 1.1.1.5

Apply pythagorean identity.

$1 + 2 \cos (x) \sin (x) - 1 = \sin (2 x)$

Step 1.1.1.6

Simplify each term.

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Step 1.1.1.6.1

Reorder $2 \cos (x)$ and $\sin (x)$ .

$1 + \sin (x) (2 \cos (x)) - 1 = \sin (2 x)$

Step 1.1.1.6.2

Reorder $\sin (x)$ and $2$ .

$1 + 2 \cdot \sin (x) \cos (x) - 1 = \sin (2 x)$

Step 1.1.1.6.3

Apply the sine double-angle identity.

$1 + \sin (2 x) - 1 = \sin (2 x)$

Step 1.1.2

Combine the opposite terms in $1 + \sin (2 x) - 1$ .

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Step 1.1.2.1

Subtract $1$ from $1$ .

$0 + \sin (2 x) = \sin (2 x)$

Step 1.1.2.2

Add $0$ and $\sin (2 x)$ .

$\sin (2 x) = \sin (2 x)$

Step 2

Move all terms containing $\sin (2 x)$ to the left side of the equation.

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Step 2.1

Subtract $\sin (2 x)$ from both sides of the equation.

$\sin (2 x) - \sin (2 x) = 0$

Step 2.2

Subtract $\sin (2 x)$ from $\sin (2 x)$ .

$0 = 0$

Step 3

Since $0 = 0$ , the equation will always be true for any value of $x$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$