Trigonometry Examples

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

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

Step 1

Simplify the left side.

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Step 1.1

Simplify

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

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

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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}

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

(\sin (x) + \cos (x)) (\sin (x) + \cos (x))

$(\sin (x) + \cos (x)) (\sin (x) + \cos (x))$ .

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

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

Step 1.1.1.2

Expand

(\sin (x) + \cos (x)) (\sin (x) + \cos (x))

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

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Step 1.1.1.2.1

Apply the distributive property.

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

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

Step 1.1.1.2.2

Apply the distributive property.

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

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

Step 1.1.1.2.3

Apply the distributive property.

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

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

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

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

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)

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

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Step 1.1.1.3.1.1.1

Raise

\sin (x)

$\sin (x)$ to the power of

1

$1$ .

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

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

Step 1.1.1.3.1.1.2

Raise

\sin (x)

$\sin (x)$ to the power of

1

$1$ .

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

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

Step 1.1.1.3.1.1.3

Use the power rule

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

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

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

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

Step 1.1.1.3.1.1.4

Add

1

$1$ and

1

$1$ .

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

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

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

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

Step 1.1.1.3.1.2

Multiply

\cos (x) \cos (x)

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

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Step 1.1.1.3.1.2.1

Raise

\cos (x)

$\cos (x)$ to the power of

1

$1$ .

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

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

Step 1.1.1.3.1.2.2

Raise

\cos (x)

$\cos (x)$ to the power of

1

$1$ .

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

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

Step 1.1.1.3.1.2.3

Use the power rule

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

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

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

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

Step 1.1.1.3.1.2.4

Add

1

$1$ and

1

$1$ .

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

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

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

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

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

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

Step 1.1.1.3.2

Reorder the factors of

\sin (x) \cos (x)

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

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

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

Step 1.1.1.3.3

Add

\cos (x) \sin (x)

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

\cos (x) \sin (x)

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

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

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

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

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

Step 1.1.1.4

Move

\cos^{2} (x)

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

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

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

Step 1.1.1.5

Apply pythagorean identity.

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

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

Step 1.1.1.6

Simplify each term.

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Step 1.1.1.6.1

Reorder

2 \cos (x)

$2 \cos (x)$ and

\sin (x)

$\sin (x)$ .

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

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

Step 1.1.1.6.2

Reorder

\sin (x)

$\sin (x)$ and

2

$2$ .

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

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

Step 1.1.1.6.3

Apply the sine double-angle identity.

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

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

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

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

Step 1.1.1.7

Apply the distributive property.

1 - 1 \cdot 1 - \sin (2 x) = - \sin (2 x)

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

Step 1.1.1.8

Multiply

- 1

$- 1$ by

1

$1$ .

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

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

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

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

Step 1.1.2

Simplify by subtracting numbers.

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Step 1.1.2.1

Subtract

1

$1$ from

1

$1$ .

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

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

Step 1.1.2.2

Subtract

\sin (2 x)

$\sin (2 x)$ from

0

$0$ .

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

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

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

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

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

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

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

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

Step 2

Move all terms containing

\sin (2 x)

$\sin (2 x)$ to the left side of the equation.

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Step 2.1

Add

\sin (2 x)

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

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

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

Step 2.2

Add

- \sin (2 x)

$- \sin (2 x)$ and

\sin (2 x)

$\sin (2 x)$ .

0 = 0

$0 = 0$

0 = 0

$0 = 0$

Step 3

Since

0 = 0

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

x

$x$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

(- \infty, \infty)

$(- \infty, \infty)$