Solve 6tan^230^circ-sqrt{3}sin60^circ-2sin45^circ

$$6 \tan ^ { 2 } 30 ^ { \circ } - \sqrt { 3 } \sin 60 ^ { \circ } - 2 \sin 45 ^ { \circ }$$

Evaluate

$$\frac{1}{2}-\sqrt{2}\approx -0.914213562$$

Solution Steps

Get the value of \tan(30) from trigonometric values table.

$$6\times \left(\frac{\sqrt{3}}{3}\right)^{2}-\sqrt{3}\sin(60)-2\sin(45)$$

To raise \frac{\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$6\times \frac{\left(\sqrt{3}\right)^{2}}{3^{2}}-\sqrt{3}\sin(60)-2\sin(45)$$

Express 6\times \frac{\left(\sqrt{3}\right)^{2}}{3^{2}} as a single fraction.

$$\frac{6\left(\sqrt{3}\right)^{2}}{3^{2}}-\sqrt{3}\sin(60)-2\sin(45)$$

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

$$\frac{6\left(\sqrt{3}\right)^{2}}{3^{2}}-\sqrt{3}\times \frac{\sqrt{3}}{2}-2\sin(45)$$

Express \sqrt{3}\times \frac{\sqrt{3}}{2} as a single fraction.

$$\frac{6\left(\sqrt{3}\right)^{2}}{3^{2}}-\frac{\sqrt{3}\sqrt{3}}{2}-2\sin(45)$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{6\left(\sqrt{3}\right)^{2}}{3^{2}}-\frac{3}{2}-2\sin(45)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3^{2} and 2 is 18. Multiply \frac{6\left(\sqrt{3}\right)^{2}}{3^{2}} times \frac{2}{2}. Multiply \frac{3}{2} times \frac{9}{9}.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}}{18}-\frac{3\times 9}{18}-2\sin(45)$$

Since \frac{2\times 6\left(\sqrt{3}\right)^{2}}{18} and \frac{3\times 9}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-2\sin(45)$$

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

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-2\times \frac{\sqrt{2}}{2}$$

Cancel out 2 and 2.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{18}{18}.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\frac{18\sqrt{2}}{18}$$

Since \frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9}{18} and \frac{18\sqrt{2}}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\times 6\left(\sqrt{3}\right)^{2}-3\times 9-18\sqrt{2}}{18}$$

Do the multiplications.

$$\frac{12\left(\sqrt{3}\right)^{2}-3\times 9}{18}-\sqrt{2}$$

The square of \sqrt{3} is 3.

$$\frac{12\times 3-3\times 9}{18}-\sqrt{2}$$

Multiply 12 and 3 to get 36.

$$\frac{36-3\times 9}{18}-\sqrt{2}$$

Multiply -3 and 9 to get -27.

$$\frac{36-27}{18}-\sqrt{2}$$

Subtract 27 from 36 to get 9.

$$\frac{9}{18}-\sqrt{2}$$

Reduce the fraction \frac{9}{18} to lowest terms by extracting and canceling out 9.

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