Solve 5sin30^{circ}+2cos^245^circ-tan^260^circ

$$5 \sin 30 ^ { \circ } + 2 \cos ^ { 2 } 45 ^ { \circ } - \tan ^ { 2 } 60 ^ { \circ }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

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

$$5\times \frac{1}{2}+2\left(\cos(45)\right)^{2}-\left(\tan(60)\right)^{2}$$

Multiply 5 and \frac{1}{2} to get \frac{5}{2}.

$$\frac{5}{2}+2\left(\cos(45)\right)^{2}-\left(\tan(60)\right)^{2}$$

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

$$\frac{5}{2}+2\times \left(\frac{\sqrt{2}}{2}\right)^{2}-\left(\tan(60)\right)^{2}$$

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

$$\frac{5}{2}+2\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-\left(\tan(60)\right)^{2}$$

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

$$\frac{5}{2}+\frac{2\left(\sqrt{2}\right)^{2}}{2^{2}}-\left(\tan(60)\right)^{2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{5}{2}+\frac{\left(\sqrt{2}\right)^{2}}{2}-\left(\tan(60)\right)^{2}$$

Since \frac{5}{2} and \frac{\left(\sqrt{2}\right)^{2}}{2} have the same denominator, add them by adding their numerators.

$$\frac{5+\left(\sqrt{2}\right)^{2}}{2}-\left(\tan(60)\right)^{2}$$

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

$$\frac{5+\left(\sqrt{2}\right)^{2}}{2}-\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{5+\left(\sqrt{2}\right)^{2}}{2}-3$$

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

$$\frac{5+\left(\sqrt{2}\right)^{2}}{2}-\frac{3\times 2}{2}$$

Since \frac{5+\left(\sqrt{2}\right)^{2}}{2} and \frac{3\times 2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5+\left(\sqrt{2}\right)^{2}-3\times 2}{2}$$

The square of \sqrt{2} is 2.

$$\frac{5+2}{2}-3$$

Add 5 and 2 to get 7.

$$\frac{7}{2}-3$$

Subtract 3 from \frac{7}{2} to get \frac{1}{2}.

$$\frac{1}{2}$$