Solve 51*sin^245^circ+cos^245=1

$$51 \cdot \sin ^ { 2 } 45 ^ { \circ } + \cos ^ { 2 } 45 = 1$$

Verify

Solution Steps

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

$$51\times \left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\cos(45)\right)^{2}=1$$

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

$$51\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}+\left(\cos(45)\right)^{2}=1$$

Express 51\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} as a single fraction.

$$\frac{51\left(\sqrt{2}\right)^{2}}{2^{2}}+\left(\cos(45)\right)^{2}=1$$

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

$$\frac{51\left(\sqrt{2}\right)^{2}}{2^{2}}+\left(\frac{\sqrt{2}}{2}\right)^{2}=1$$

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

$$\frac{51\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}=1$$

Since \frac{51\left(\sqrt{2}\right)^{2}}{2^{2}} and \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{51\left(\sqrt{2}\right)^{2}+\left(\sqrt{2}\right)^{2}}{2^{2}}=1$$

The square of \sqrt{2} is 2.

$$\frac{51\times 2+\left(\sqrt{2}\right)^{2}}{2^{2}}=1$$

Multiply 51 and 2 to get 102.

$$\frac{102+\left(\sqrt{2}\right)^{2}}{2^{2}}=1$$

The square of \sqrt{2} is 2.

$$\frac{102+2}{2^{2}}=1$$

Add 102 and 2 to get 104.

$$\frac{104}{2^{2}}=1$$

Calculate 2 to the power of 2 and get 4.

$$\frac{104}{4}=1$$

Divide 104 by 4 to get 26.

$$26=1$$

Compare 26 and 1.

$$\text{false}$$