Solve 1/2-frac{{left(frac{2+sqrt{3}}{2}right)}^2}{2}

$$\frac{ 1 }{ 2 } - \frac{ { \left( \frac{ 2+ \sqrt{ 3 } }{ 2 } \right) }^{ 2 } }{ 2 }$$

Evaluate

$$-\frac{\sqrt{3}}{2}-\frac{3}{8}\approx -1.241025404$$

Short Solution Steps

To raise \frac{2+\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1}{2}-\frac{\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}}}{2}$$

Express \frac{\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}}}{2} as a single fraction.

$$\frac{1}{2}-\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}\times 2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

$$\frac{1}{2}-\frac{4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2^{2}\times 2}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}-\frac{4+4\sqrt{3}+3}{2^{2}\times 2}$$

Add 4 and 3 to get 7.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{2^{2}\times 2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{2^{3}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{8}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 8 is 8. Multiply \frac{1}{2} times \frac{4}{4}.

$$\frac{4}{8}-\frac{7+4\sqrt{3}}{8}$$

Since \frac{4}{8} and \frac{7+4\sqrt{3}}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-\left(7+4\sqrt{3}\right)}{8}$$

Do the multiplications in 4-\left(7+4\sqrt{3}\right).

$$\frac{4-7-4\sqrt{3}}{8}$$

Do the calculations in 4-7-4\sqrt{3}.

$$\frac{-3-4\sqrt{3}}{8}$$

Expand

$$-\frac{\sqrt{3}}{2} - \frac{3}{8} = -1.241025404$$

Short Solution Steps

To raise \frac{2+\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{1}{2}-\frac{\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}}}{2}$$

Express \frac{\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}}}{2} as a single fraction.

$$\frac{1}{2}-\frac{\left(2+\sqrt{3}\right)^{2}}{2^{2}\times 2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

$$\frac{1}{2}-\frac{4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2^{2}\times 2}$$

The square of \sqrt{3} is 3.

$$\frac{1}{2}-\frac{4+4\sqrt{3}+3}{2^{2}\times 2}$$

Add 4 and 3 to get 7.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{2^{2}\times 2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{2^{3}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1}{2}-\frac{7+4\sqrt{3}}{8}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 8 is 8. Multiply \frac{1}{2} times \frac{4}{4}.

$$\frac{4}{8}-\frac{7+4\sqrt{3}}{8}$$

Since \frac{4}{8} and \frac{7+4\sqrt{3}}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-\left(7+4\sqrt{3}\right)}{8}$$

Do the multiplications in 4-\left(7+4\sqrt{3}\right).

$$\frac{4-7-4\sqrt{3}}{8}$$

Do the calculations in 4-7-4\sqrt{3}.

$$\frac{-3-4\sqrt{3}}{8}$$