Convert to Rectangular ( square root of 3(cos((5pi)/12)+isin((5pi)/12)))^4

${(\sqrt{3} (\cos (\frac{5 π}{12}) + i \sin (\frac{5 π}{12})))}^{4}$

Step 1

Simplify each term.

The exact value of $\cos (\frac{5 π}{12})$ is $\frac{\sqrt{6} - \sqrt{2}}{4}$ .

Split $\frac{5 π}{12}$ into two angles where the values of the six trigonometric functions are known.

${(\sqrt{3} (\cos (\frac{π}{6} + \frac{π}{4}) + i \sin (\frac{5 π}{12})))}^{4}$

Apply the sum of angles identity $\cos (x + y) = \cos (x) \cos (y) - \sin (x) \sin (y)$ .

${(\sqrt{3} (\cos (\frac{π}{6}) \cos (\frac{π}{4}) - \sin (\frac{π}{6}) \sin (\frac{π}{4}) + i \sin (\frac{5 π}{12})))}^{4}$

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{3}}{2} \cos (\frac{π}{4}) - \sin (\frac{π}{6}) \sin (\frac{π}{4}) + i \sin (\frac{5 π}{12})))}^{4}$

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \sin (\frac{π}{6}) \sin (\frac{π}{4}) + i \sin (\frac{5 π}{12})))}^{4}$

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

${(\sqrt{3} (\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \sin (\frac{π}{4}) + i \sin (\frac{5 π}{12})))}^{4}$

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i \sin (\frac{5 π}{12})))}^{4}$

Simplify $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ .

Simplify each term.

Multiply $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ .

Multiply $\frac{\sqrt{3}}{2}$ by $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{3} \sqrt{2}}{2 \cdot 2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i \sin (\frac{5 π}{12})))}^{4}$

Combine using the product rule for radicals.

${(\sqrt{3} (\frac{\sqrt{3 \cdot 2}}{2 \cdot 2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i \sin (\frac{5 π}{12})))}^{4}$

Multiply $3$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6}}{2 \cdot 2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i \sin (\frac{5 π}{12})))}^{4}$

Multiply $2$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6}}{4} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + i \sin (\frac{5 π}{12})))}^{4}$

Multiply $- \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ .

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{1}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2 \cdot 2} + i \sin (\frac{5 π}{12})))}^{4}$

Multiply $2$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + i \sin (\frac{5 π}{12})))}^{4}$

Combine the numerators over the common denominator.

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i \sin (\frac{5 π}{12})))}^{4}$

The exact value of $\sin (\frac{5 π}{12})$ is $\frac{\sqrt{2} + \sqrt{6}}{4}$ .

Split $\frac{5 π}{12}$ into two angles where the values of the six trigonometric functions are known.

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i \sin (\frac{π}{6} + \frac{π}{4})))}^{4}$

Apply the sum of angles identity.

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\sin (\frac{π}{6}) \cos (\frac{π}{4}) + \cos (\frac{π}{6}) \sin (\frac{π}{4}))))}^{4}$

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{1}{2} \cos (\frac{π}{4}) + \cos (\frac{π}{6}) \sin (\frac{π}{4}))))}^{4}$

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \cos (\frac{π}{6}) \sin (\frac{π}{4}))))}^{4}$

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \sin (\frac{π}{4}))))}^{4}$

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2})))}^{4}$

Simplify $\frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ .

Simplify each term.

Multiply $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{2 \cdot 2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2})))}^{4}$

Multiply $2$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{4} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2})))}^{4}$

Multiply $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ .

Multiply $\frac{\sqrt{3}}{2}$ by $\frac{\sqrt{2}}{2}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{4} + \frac{\sqrt{3} \sqrt{2}}{2 \cdot 2})))}^{4}$

Combine using the product rule for radicals.

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{4} + \frac{\sqrt{3 \cdot 2}}{2 \cdot 2})))}^{4}$

Multiply $3$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{2 \cdot 2})))}^{4}$

Multiply $2$ by $2$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i (\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4})))}^{4}$

Combine the numerators over the common denominator.

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{2} + \sqrt{6}}{4}))}^{4}$

Combine $i$ and $\frac{\sqrt{2} + \sqrt{6}}{4}$ .

${(\sqrt{3} (\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{i (\sqrt{2} + \sqrt{6})}{4}))}^{4}$

Step 2

Combine fractions.

Combine the numerators over the common denominator.

${(\sqrt{3} \frac{\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6}}{4})}^{4}$

Combine $\sqrt{3}$ and $\frac{\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6}}{4}$ .

${(\frac{\sqrt{3} (\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}{4})}^{4}$

Step 3

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

Apply the product rule to $\frac{\sqrt{3} (\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}{4}$ .

$\frac{{(\sqrt{3} (\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6}))}^{4}}{4^{4}}$

Apply the product rule to $\sqrt{3} (\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})$ .

$\frac{{\sqrt{3}}^{4} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Step 4

Simplify the numerator.

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

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

$\frac{{(3^{\frac{1}{2}})}^{4} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

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

$\frac{3^{\frac{1}{2} \cdot 4} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

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

$\frac{3^{\frac{4}{2}} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

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

Factor $2$ out of $4$ .

$\frac{3^{\frac{2 \cdot 2}{2}} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{3^{\frac{2 \cdot 2}{2 (1)}} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Cancel the common factor.

$\frac{3^{\frac{2 \cdot 2}{2 \cdot 1}} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Rewrite the expression.

$\frac{3^{\frac{2}{1}} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Divide $2$ by $1$ .

$\frac{3^{2} {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

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

$\frac{9 {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{4^{4}}$

Step 5

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

$\frac{9 {(\sqrt{6} - \sqrt{2} + i \sqrt{2} + i \sqrt{6})}^{4}}{256}$

Convert to Rectangular ( square root of 3(cos((5pi)/12)+isin((5pi)/12)))^4

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