Solve =[1/sqrt{2}*frac{sqrt{3}}{2}+1/sqrt{2}*1/2]*[1/sqrt{2}*frac{sqrt{3}}{2}+frac{1}{sqrt{2}}*frac{1}{2}]

$$= [ \frac { 1 } { \sqrt { 2 } } \times \frac { \sqrt { 3 } } { 2 } + \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { 2 } ] \times [ \frac { 1 } { \sqrt { 2 } } \times \frac { \sqrt { 3 } } { 2 } + \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { 2 } ]$$

Evaluate

$$\frac{\sqrt{3}+2}{4}\approx 0.933012702$$

Short Solution Steps

Multiply \frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2} and \frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2} to get \left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}\right)^{2}.

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

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply \frac{\sqrt{2}}{2} times \frac{\sqrt{3}}{2} by multiplying numerator times numerator and denominator times denominator.

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

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply \frac{\sqrt{2}}{2} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

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

Since \frac{\sqrt{2}\sqrt{3}}{2\times 2} and \frac{\sqrt{2}}{2\times 2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \sqrt{2}\sqrt{3}+\sqrt{2}.

$$\left(\frac{\sqrt{6}+\sqrt{2}}{2\times 2}\right)^{2}$$

To raise \frac{\sqrt{6}+\sqrt{2}}{2\times 2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{6}+\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+\sqrt{2}\right)^{2}.

$$\frac{\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{6+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

Multiply 2 and 2 to get 4.

$$\frac{6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}}{\left(2\times 2\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{6+4\sqrt{3}+2}{\left(2\times 2\right)^{2}}$$

Add 6 and 2 to get 8.

$$\frac{8+4\sqrt{3}}{\left(2\times 2\right)^{2}}$$

Multiply 2 and 2 to get 4.

$$\frac{8+4\sqrt{3}}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{8+4\sqrt{3}}{16}$$

Factor

$$\frac{\sqrt{3} + 2}{4} = 0.9330127018922193$$