Solve sqrt{18}-(pi-1)^0-2cos45^circ+(1/2)^-2

$$\sqrt { 18 } - ( \pi - 1 ) ^ { 0 } - 2 \cos 45 ^ { \circ } + ( \frac { 1 } { 2 } ) ^ { - 2 }$$

Evaluate

$$2\sqrt{2}+3\approx 5.828427125$$

Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$3\sqrt{2}-\left(\pi -1\right)^{0}-2\cos(45)+\left(\frac{1}{2}\right)^{-2}$$

Calculate \pi -1 to the power of 0 and get 1.

$$3\sqrt{2}-1-2\cos(45)+\left(\frac{1}{2}\right)^{-2}$$

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

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

Cancel out 2 and 2.

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

Combine 3\sqrt{2} and -\sqrt{2} to get 2\sqrt{2}.

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

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

$$2\sqrt{2}-1+4$$

Add -1 and 4 to get 3.

$$2\sqrt{2}+3$$