Trigonometry Examples

${(- 3 + 3 i \sqrt{3})}^{5}$

Step 1

Use the Binomial Theorem.

${(- 3)}^{5} + 5 {(- 3)}^{4} (3 i \sqrt{3}) + 10 {(- 3)}^{3} {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2

Simplify terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

Raise $- 3$ to the power of $5$ .

$- 243 + 5 {(- 3)}^{4} (3 i \sqrt{3}) + 10 {(- 3)}^{3} {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.2

Raise $- 3$ to the power of $4$ .

$- 243 + 5 \cdot 81 (3 i \sqrt{3}) + 10 {(- 3)}^{3} {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.3

Multiply $5$ by $81$ .

$- 243 + 405 (3 i \sqrt{3}) + 10 {(- 3)}^{3} {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.4

Multiply $3$ by $405$ .

$- 243 + 1215 (i \sqrt{3}) + 10 {(- 3)}^{3} {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.5

Raise $- 3$ to the power of $3$ .

$- 243 + 1215 i \sqrt{3} + 10 \cdot - 27 {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.6

Multiply $10$ by $- 27$ .

$- 243 + 1215 i \sqrt{3} - 270 {(3 i \sqrt{3})}^{2} + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.7

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.1.7.1

Apply the product rule to $3 i \sqrt{3}$ .

$- 243 + 1215 i \sqrt{3} - 270 ({(3 i)}^{2} {\sqrt{3}}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.7.2

Apply the product rule to $3 i$ .

$- 243 + 1215 i \sqrt{3} - 270 (3^{2} i^{2} {\sqrt{3}}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.8

Raise $3$ to the power of $2$ .

$- 243 + 1215 i \sqrt{3} - 270 (9 i^{2} {\sqrt{3}}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.9

Rewrite $i^{2}$ as $- 1$ .

$- 243 + 1215 i \sqrt{3} - 270 (9 \cdot - 1 {\sqrt{3}}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.10

Multiply $9$ by $- 1$ .

$- 243 + 1215 i \sqrt{3} - 270 (- 9 {\sqrt{3}}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 2.1.11.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$- 243 + 1215 i \sqrt{3} - 270 (- 9 {(3^{\frac{1}{2}})}^{2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 243 + 1215 i \sqrt{3} - 270 (- 9 \cdot 3^{\frac{1}{2} \cdot 2}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11.3

Combine $\frac{1}{2}$ and $2$ .

$- 243 + 1215 i \sqrt{3} - 270 (- 9 \cdot 3^{\frac{2}{2}}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11.4

Cancel the common factor of $2$ .

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Step 2.1.11.4.1

Cancel the common factor.

$- 243 + 1215 i \sqrt{3} - 270 (- 9 \cdot 3^{\frac{2}{2}}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11.4.2

Rewrite the expression.

$- 243 + 1215 i \sqrt{3} - 270 (- 9 \cdot 3^{1}) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.11.5

Evaluate the exponent.

$- 243 + 1215 i \sqrt{3} - 270 (- 9 \cdot 3) + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.12

Multiply $- 270 (- 9 \cdot 3)$ .

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Step 2.1.12.1

Multiply $- 9$ by $3$ .

$- 243 + 1215 i \sqrt{3} - 270 \cdot - 27 + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.12.2

Multiply $- 270$ by $- 27$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 10 {(- 3)}^{2} {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.13

Raise $- 3$ to the power of $2$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 10 \cdot 9 {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.14

Multiply $10$ by $9$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 {(3 i \sqrt{3})}^{3} + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.15

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.1.15.1

Apply the product rule to $3 i \sqrt{3}$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 ({(3 i)}^{3} {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.15.2

Apply the product rule to $3 i$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (3^{3} i^{3} {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.16

Raise $3$ to the power of $3$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (27 i^{3} {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.17

Factor out $i^{2}$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (27 (i^{2} \cdot i) {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.18

Rewrite $i^{2}$ as $- 1$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (27 (- 1 \cdot i) {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.19

Rewrite $- 1 i$ as $- i$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (27 (- i) {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.20

Multiply $- 1$ by $27$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i {\sqrt{3}}^{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.21

Rewrite ${\sqrt{3}}^{3}$ as $\sqrt{3^{3}}$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i \sqrt{3^{3}}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.22

Raise $3$ to the power of $3$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i \sqrt{27}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.23

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 2.1.23.1

Factor $9$ out of $27$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i \sqrt{9 (3)}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.23.2

Rewrite $9$ as $3^{2}$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i \sqrt{3^{2} \cdot 3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.24

Pull terms out from under the radical.

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 27 i (3 \sqrt{3})) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.25

Multiply $3$ by $- 27$ .

$- 243 + 1215 i \sqrt{3} + 7290 + 90 (- 81 i \sqrt{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.26

Multiply $- 81$ by $90$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 (i \sqrt{3}) + 5 \cdot - 3 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.27

Multiply $5$ by $- 3$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 {(3 i \sqrt{3})}^{4} + {(3 i \sqrt{3})}^{5}$

Step 2.1.28

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.1.28.1

Apply the product rule to $3 i \sqrt{3}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 ({(3 i)}^{4} {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.28.2

Apply the product rule to $3 i$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (3^{4} i^{4} {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.29

Raise $3$ to the power of $4$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 i^{4} {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.30

Rewrite $i^{4}$ as $1$ .

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Step 2.1.30.1

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 {(i^{2})}^{2} {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.30.2

Rewrite $i^{2}$ as $- 1$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 {(- 1)}^{2} {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.30.3

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

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 1 {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.31

Multiply $81$ by $1$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 {\sqrt{3}}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

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Step 2.1.32.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 {(3^{\frac{1}{2}})}^{4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{1}{2} \cdot 4}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.3

Combine $\frac{1}{2}$ and $4$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{4}{2}}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.4

Cancel the common factor of $4$ and $2$ .

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Step 2.1.32.4.1

Factor $2$ out of $4$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{2 \cdot 2}{2}}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.4.2

Cancel the common factors.

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Step 2.1.32.4.2.1

Factor $2$ out of $2$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{2 \cdot 2}{2 (1)}}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.4.2.2

Cancel the common factor.

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{2 \cdot 2}{2 \cdot 1}}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.4.2.3

Rewrite the expression.

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{\frac{2}{1}}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.32.4.2.4

Divide $2$ by $1$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 3^{2}) + {(3 i \sqrt{3})}^{5}$

Step 2.1.33

Raise $3$ to the power of $2$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 (81 \cdot 9) + {(3 i \sqrt{3})}^{5}$

Step 2.1.34

Multiply $- 15 (81 \cdot 9)$ .

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Step 2.1.34.1

Multiply $81$ by $9$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 15 \cdot 729 + {(3 i \sqrt{3})}^{5}$

Step 2.1.34.2

Multiply $- 15$ by $729$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + {(3 i \sqrt{3})}^{5}$

Step 2.1.35

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 2.1.35.1

Apply the product rule to $3 i \sqrt{3}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + {(3 i)}^{5} {\sqrt{3}}^{5}$

Step 2.1.35.2

Apply the product rule to $3 i$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 3^{5} i^{5} {\sqrt{3}}^{5}$

Step 2.1.36

Raise $3$ to the power of $5$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i^{5} {\sqrt{3}}^{5}$

Step 2.1.37

Factor out $i^{4}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 (i^{4} i) {\sqrt{3}}^{5}$

Step 2.1.38

Rewrite $i^{4}$ as $1$ .

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Step 2.1.38.1

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 ({(i^{2})}^{2} i) {\sqrt{3}}^{5}$

Step 2.1.38.2

Rewrite $i^{2}$ as $- 1$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 ({(- 1)}^{2} i) {\sqrt{3}}^{5}$

Step 2.1.38.3

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

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 (1 i) {\sqrt{3}}^{5}$

Step 2.1.39

Multiply $i$ by $1$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i {\sqrt{3}}^{5}$

Step 2.1.40

Rewrite ${\sqrt{3}}^{5}$ as $\sqrt{3^{5}}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i \sqrt{3^{5}}$

Step 2.1.41

Raise $3$ to the power of $5$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i \sqrt{243}$

Step 2.1.42

Rewrite $243$ as $9^{2} \cdot 3$ .

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Step 2.1.42.1

Factor $81$ out of $243$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i \sqrt{81 (3)}$

Step 2.1.42.2

Rewrite $81$ as $9^{2}$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i \sqrt{9^{2} \cdot 3}$

Step 2.1.43

Pull terms out from under the radical.

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 243 i (9 \sqrt{3})$

Step 2.1.44

Multiply $9$ by $243$ .

$- 243 + 1215 i \sqrt{3} + 7290 - 7290 i \sqrt{3} - 10935 + 2187 i \sqrt{3}$

Step 2.2

Simplify by adding terms.

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Step 2.2.1

Add $- 243$ and $7290$ .

$1215 i \sqrt{3} + 7047 - 7290 i \sqrt{3} - 10935 + 2187 i \sqrt{3}$

Step 2.2.2

Subtract $7290 i \sqrt{3}$ from $1215 i \sqrt{3}$ .

$- 6075 i \sqrt{3} + 7047 - 10935 + 2187 i \sqrt{3}$

Step 2.2.3

Add $- 6075 i \sqrt{3}$ and $2187 i \sqrt{3}$ .

$- 3888 i \sqrt{3} + 7047 - 10935$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Subtract $10935$ from $7047$ .

$- 3888 i \sqrt{3} - 3888$

Step 2.2.4.2

Reorder $- 3888 i \sqrt{3}$ and $- 3888$ .

$- 3888 - 3888 i \sqrt{3}$