Solve -sin60^circ+2sin30^circ+3sin45^circ/sqrt{2}

$$- \frac { \sin 60 ^ { \circ } + 2 \sin 30 ^ { \circ } + 3 \sin 45 ^ { \circ } } { \sqrt { 2 } }$$

Evaluate

$$-\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{4}-\frac{3}{2}\approx -2.819479217$$

Short Solution Steps

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

$$-\frac{\frac{\sqrt{3}}{2}+2\sin(30)+3\sin(45)}{\sqrt{2}}$$

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

$$-\frac{\frac{\sqrt{3}}{2}+2\times \frac{1}{2}+3\sin(45)}{\sqrt{2}}$$

Multiply 2 and \frac{1}{2} to get 1.

$$-\frac{\frac{\sqrt{3}}{2}+1+3\sin(45)}{\sqrt{2}}$$

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

$$-\frac{\frac{\sqrt{3}}{2}+1+3\times \frac{\sqrt{2}}{2}}{\sqrt{2}}$$

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

$$-\frac{\frac{\sqrt{3}}{2}+1+\frac{3\sqrt{2}}{2}}{\sqrt{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{\sqrt{3}}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

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

Since \frac{\sqrt{3}+2}{2} and \frac{3\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

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

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

Rationalize the denominator of \frac{\sqrt{3}+2+3\sqrt{2}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

Use the distributive property to multiply \sqrt{3}+2+3\sqrt{2} by \sqrt{2}.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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