Precalculus Examples

$\sin (\arccos (\frac{\sqrt{3}}{2} + \arctan (1)))$

Step 1

Combine the numerators over the common denominator.

$\sqrt{\frac{1 - π \sqrt{3}}{4} - \frac{π^{2}}{16}}$

Step 2

To write $\frac{1 - π \sqrt{3}}{4}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\sqrt{\frac{1 - π \sqrt{3}}{4} \cdot \frac{4}{4} - \frac{π^{2}}{16}}$

Step 3

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1

Multiply $\frac{1 - π \sqrt{3}}{4}$ by $\frac{4}{4}$ .

$\sqrt{\frac{(1 - π \sqrt{3}) \cdot 4}{4 \cdot 4} - \frac{π^{2}}{16}}$

Step 3.2

Multiply $4$ by $4$ .

$\sqrt{\frac{(1 - π \sqrt{3}) \cdot 4}{16} - \frac{π^{2}}{16}}$

Step 4

Combine the numerators over the common denominator.

$\sqrt{\frac{(1 - π \sqrt{3}) \cdot 4 - π^{2}}{16}}$

Step 5

Simplify the numerator.

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Step 5.1

Apply the distributive property.

$\sqrt{\frac{1 \cdot 4 - π \sqrt{3} \cdot 4 - π^{2}}{16}}$

Step 5.2

Multiply $4$ by $1$ .

$\sqrt{\frac{4 - π \sqrt{3} \cdot 4 - π^{2}}{16}}$

Step 5.3

Multiply $4$ by $- 1$ .

$\sqrt{\frac{4 - 4 π \sqrt{3} - π^{2}}{16}}$

Step 6

Rewrite $\sqrt{\frac{4 - 4 π \sqrt{3} - π^{2}}{16}}$ as $\frac{\sqrt{4 - 4 π \sqrt{3} - π^{2}}}{\sqrt{16}}$ .

$\frac{\sqrt{4 - 4 π \sqrt{3} - π^{2}}}{\sqrt{16}}$

Step 7

Simplify the denominator.

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Step 7.1

Rewrite $16$ as $4^{2}$ .

$\frac{\sqrt{4 - 4 π \sqrt{3} - π^{2}}}{\sqrt{4^{2}}}$

Step 7.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{\sqrt{4 - 4 π \sqrt{3} - π^{2}}}{4}$