Solve (sinpi/4cospi/6-sinpi/6cospi/4)^2

$$( \sin \frac { \pi } { 4 } \cos \frac { \pi } { 6 } - \sin \frac { \pi } { 6 } \cos \frac { \pi } { 4 } ) ^ { 2 }$$

Evaluate

$$-\frac{\sqrt{3}}{4}+\frac{1}{2}\approx 0.066987298$$

Short Solution Steps

Get the value of \sin(\frac{\pi }{4}) from trigonometric values table.

$$\left(\frac{\sqrt{2}}{2}\cos(\frac{\pi }{6})-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

Get the value of \cos(\frac{\pi }{6}) from trigonometric values table.

$$\left(\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

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

$$\left(\frac{\sqrt{2}\sqrt{3}}{2\times 2}-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

Get the value of \sin(\frac{\pi }{6}) from trigonometric values table.

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

Get the value of \cos(\frac{\pi }{4}) from trigonometric values table.

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

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

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

Multiply 2 and 2 to get 4.

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

To add or subtract expressions, expand them to make their denominators the same. Expand 2\times 2.

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

Since \frac{\sqrt{2}\sqrt{3}}{4} and \frac{\sqrt{2}}{4} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^{2}$$

To raise \frac{\sqrt{6}-\sqrt{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{6}-\sqrt{2}\right)^{2}}{4^{2}}$$

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

Calculate 4 to the power of 2 and get 16.

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

Expand

$$-\frac{\sqrt{3}}{4} + \frac{1}{2} = 0.066987298$$

Short Solution Steps

Get the value of \sin(\frac{\pi }{4}) from trigonometric values table.

$$\left(\frac{\sqrt{2}}{2}\cos(\frac{\pi }{6})-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

Get the value of \cos(\frac{\pi }{6}) from trigonometric values table.

$$\left(\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

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

$$\left(\frac{\sqrt{2}\sqrt{3}}{2\times 2}-\sin(\frac{\pi }{6})\cos(\frac{\pi }{4})\right)^{2}$$

Get the value of \sin(\frac{\pi }{6}) from trigonometric values table.

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

Get the value of \cos(\frac{\pi }{4}) from trigonometric values table.

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

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

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

Multiply 2 and 2 to get 4.

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

To add or subtract expressions, expand them to make their denominators the same. Expand 2\times 2.

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

Since \frac{\sqrt{2}\sqrt{3}}{4} and \frac{\sqrt{2}}{4} have the same denominator, subtract them by subtracting their numerators.

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

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

$$\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^{2}$$

To raise \frac{\sqrt{6}-\sqrt{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{6}-\sqrt{2}\right)^{2}}{4^{2}}$$

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

Calculate 4 to the power of 2 and get 16.

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