Solve sin^260^circ-3*tan45^circ

$$\sin ^ { 2 } 60 ^ { \circ } - 3 \cdot \tan 45 ^ { \circ }$$

Evaluate

$$-\frac{9}{4}=-2.25$$

Solution Steps

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

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

To raise \frac{\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

Multiply 3 and 1 to get 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2^{2}}{2^{2}}.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{3}{2^{2}}-3$$

Calculate 2 to the power of 2 and get 4.

$$\frac{3}{4}-3$$

Subtract 3 from \frac{3}{4} to get -\frac{9}{4}.

$$-\frac{9}{4}$$