Verify the Identity cos(x)^4-sin(x)^4=1-2sin(x)^2

$\cos^{4} (x) - \sin^{4} (x) = 1 - 2 \sin^{2} (x)$

Start on the left side.

$\cos^{4} (x) - \sin^{4} (x)$

Factor.

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Rewrite $\cos^{4} (x)$ as ${(\cos^{2} (x))}^{2}$ .

${(\cos^{2} (x))}^{2} - \sin^{4} (x)$

Rewrite $\sin^{4} (x)$ as ${(\sin^{2} (x))}^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos^{2} (x)$ and $b = \sin^{2} (x)$ .

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

Simplify.

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Apply pythagorean identity.

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = \sin (x)$ .

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

Apply the distributive property.

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

Simplify.

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Simplify each term.

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Apply the distributive property.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Apply the distributive property.

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Simplify each term.

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Move $- 1$ to the left of $\cos (x)$ .

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

Rewrite $- 1 \cos (x)$ as $- \cos (x)$ .

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

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

${\cos (x)}^{2} + 0 - {\sin (x)}^{2}$

Add ${\cos (x)}^{2}$ and $0$ .

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

Apply Pythagorean identity in reverse.

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

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

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

Because the two sides have been shown to be equivalent, the equation is an identity.

$\cos^{4} (x) - \sin^{4} (x) = 1 - 2 \sin^{2} (x)$ is an identity

Verify the Identity cos(x)^4-sin(x)^4=1-2sin(x)^2

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