Precalculus Examples

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

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(\sin (x))}^{2 - k} \cdot {(- \cos (x))}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(\sin (x))}^{2 - 0} \cdot {(- \cos (x))}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(\sin (x))}^{2 - 1} \cdot {(- \cos (x))}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(\sin (x))}^{2 - 2} \cdot {(- \cos (x))}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

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Step 4.1

Simplify terms.

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

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

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

Step 4.1.1.2

Apply the product rule to $- \cos (x)$ .

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

Step 4.1.1.3

Rewrite using the commutative property of multiplication.

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

Step 4.1.1.4

Anything raised to $0$ is $1$ .

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

Step 4.1.1.5

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

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

Step 4.1.1.6

Anything raised to $0$ is $1$ .

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

Step 4.1.1.7

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

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

Step 4.1.1.8

Simplify.

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

Step 4.1.1.9

Simplify.

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

Step 4.1.1.10

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

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

Step 4.1.1.11

Add parentheses.

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

Step 4.1.1.12

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

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

Step 4.1.1.13

Apply the sine double-angle identity.

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

Step 4.1.1.14

Multiply $\sin^{0} (x)$ by $1$ .

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

Step 4.1.1.15

Anything raised to $0$ is $1$ .

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

Step 4.1.1.16

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

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

Step 4.1.1.17

Apply the product rule to $- \cos (x)$ .

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

Step 4.1.1.18

Raise $- 1$ to the power of $2$ .

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

Step 4.1.1.19

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

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

Step 4.1.2

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

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

Step 4.2

Apply pythagorean identity.

$1 - \sin (2 x)$