Solve {left(-1/2right)}^2+1+2-sqrt{3}+2sin(60^circ)-sqrt{8}

$${ \left(- \frac{ 1 }{ 2 } \right) }^{ 2 } +1+2- \sqrt{ 3 } +2 \sin ( 60 ^ { \circ } ) - \sqrt{ 8 }$$

Evaluate

$$\frac{13}{4}-2\sqrt{2}\approx 0.421572875$$

Solution Steps

Calculate -\frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}+1+2-\sqrt{3}+2\sin(60)-\sqrt{8}$$

Add \frac{1}{4} and 1 to get \frac{5}{4}.

$$\frac{5}{4}+2-\sqrt{3}+2\sin(60)-\sqrt{8}$$

Add \frac{5}{4} and 2 to get \frac{13}{4}.

$$\frac{13}{4}-\sqrt{3}+2\sin(60)-\sqrt{8}$$

Get the value of \sin(60) from trigonometric values table.

$$\frac{13}{4}-\sqrt{3}+2\times \frac{\sqrt{3}}{2}-\sqrt{8}$$

Cancel out 2 and 2.

$$\frac{13}{4}-\sqrt{3}+\sqrt{3}-\sqrt{8}$$

Combine -\sqrt{3} and \sqrt{3} to get 0.

$$\frac{13}{4}-\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{13}{4}-2\sqrt{2}$$