Solve cos30^circ/sin60^circ+cos45^circ

$$\frac { \cos 30 ^ { \circ } } { \sin 60 ^ { \circ } + \cos 45 ^ { \circ } }$$

Evaluate

$$3-\sqrt{6}\approx 0.550510257$$

Solution Steps

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

$$\frac{\frac{\sqrt{3}}{2}}{\sin(60)+\cos(45)}$$

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

$$\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}+\cos(45)}$$

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

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

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

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

Divide \frac{\sqrt{3}}{2} by \frac{\sqrt{3}+\sqrt{2}}{2} by multiplying \frac{\sqrt{3}}{2} by the reciprocal of \frac{\sqrt{3}+\sqrt{2}}{2}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

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

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

The square of \sqrt{3} is 3.

$$-\left(\sqrt{6}-3\right)$$

To find the opposite of \sqrt{6}-3, find the opposite of each term.

$$-\sqrt{6}+3$$