Evaluate square root of (1-cos((7pi)/6))/2

$\sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$\sqrt{\frac{1 - - \cos (\frac{π}{6})}{2}}$

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

$\sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{2}}$

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

Multiply $- 1$ by $- 1$ .

$\sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}}$

Multiply $\frac{\sqrt{3}}{2}$ by $1$ .

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

Simplify the numerator.

Write $1$ as a fraction with a common denominator.

$\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}}$

Combine the numerators over the common denominator.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}}$

Multiply the numerator by the reciprocal of the denominator.

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

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

Multiply $\frac{2 + \sqrt{3}}{2}$ by $\frac{1}{2}$ .

$\sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}}$

Multiply $2$ by $2$ .

$\sqrt{\frac{2 + \sqrt{3}}{4}}$

Rewrite $\sqrt{\frac{2 + \sqrt{3}}{4}}$ as $\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$ .

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$

Simplify the denominator.

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

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2^{2}}}$

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

$\frac{\sqrt{2 + \sqrt{3}}}{2}$

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{2 + \sqrt{3}}}{2}$

Decimal Form:

$0.96592582 \dots$

Evaluate square root of (1-cos((7pi)/6))/2

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