Solve |-1/2|+(-1)^2019+2^-1-(2-sqrt{2})^0+2cos45^circ

$$| - \frac { 1 } { 2 } | + ( - 1 ) ^ { 2019 } + 2 ^ { - 1 } - ( 2 - \sqrt { 2 } ) ^ { 0 } + 2 \cos 45 ^ { \circ }$$

Evaluate

$$\sqrt{2}-1\approx 0.414213562$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{1}{2} is \frac{1}{2}.

$$\frac{1}{2}+\left(-1\right)^{2019}+2^{-1}-\left(2-\sqrt{2}\right)^{0}+2\cos(45)$$

Calculate -1 to the power of 2019 and get -1.

$$\frac{1}{2}-1+2^{-1}-\left(2-\sqrt{2}\right)^{0}+2\cos(45)$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$-\frac{1}{2}+2^{-1}-\left(2-\sqrt{2}\right)^{0}+2\cos(45)$$

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

$$-\frac{1}{2}+\frac{1}{2}-\left(2-\sqrt{2}\right)^{0}+2\cos(45)$$

Add -\frac{1}{2} and \frac{1}{2} to get 0.

$$0-\left(2-\sqrt{2}\right)^{0}+2\cos(45)$$

Calculate 2-\sqrt{2} to the power of 0 and get 1.

$$0-1+2\cos(45)$$

Subtract 1 from 0 to get -1.

$$-1+2\cos(45)$$

Get the value of \cos(45) from trigonometric values table.

$$-1+2\times \frac{\sqrt{2}}{2}$$

Cancel out 2 and 2.

$$-1+\sqrt{2}$$