Solve cos90^{circ}=1-2sin^245^circ=2cos^245^circ-1

$$\cos 90 ^ { \circ } = 1 - 2 \sin ^ { 2 } 45 ^ { \circ } = 2 \cos ^ { 2 } 45 ^ { \circ } - 1$$

Verify

Solution Steps

Get the value of \cos(90) from trigonometric values table.

$$0=1-2\left(\sin(45)\right)^{2}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

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

$$0=1-2\times \left(\frac{\sqrt{2}}{2}\right)^{2}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

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

$$0=1-2\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

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

$$0=1-\frac{2\left(\sqrt{2}\right)^{2}}{2^{2}}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

Cancel out 2 in both numerator and denominator.

$$0=1-\frac{\left(\sqrt{2}\right)^{2}}{2}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

The square of \sqrt{2} is 2.

$$0=1-\frac{2}{2}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

Divide 2 by 2 to get 1.

$$0=1-1\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

Subtract 1 from 1 to get 0.

$$0=0\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

Compare 0 and 0.

$$\text{true}\text{ and }1-2\left(\sin(45)\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

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

$$\text{true}\text{ and }1-2\times \left(\frac{\sqrt{2}}{2}\right)^{2}=2\left(\cos(45)\right)^{2}-1$$

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

$$\text{true}\text{ and }1-2\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}=2\left(\cos(45)\right)^{2}-1$$

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

$$\text{true}\text{ and }1-\frac{2\left(\sqrt{2}\right)^{2}}{2^{2}}=2\left(\cos(45)\right)^{2}-1$$

Cancel out 2 in both numerator and denominator.

$$\text{true}\text{ and }1-\frac{\left(\sqrt{2}\right)^{2}}{2}=2\left(\cos(45)\right)^{2}-1$$

The square of \sqrt{2} is 2.

$$\text{true}\text{ and }1-\frac{2}{2}=2\left(\cos(45)\right)^{2}-1$$

Divide 2 by 2 to get 1.

$$\text{true}\text{ and }1-1=2\left(\cos(45)\right)^{2}-1$$

Subtract 1 from 1 to get 0.

$$\text{true}\text{ and }0=2\left(\cos(45)\right)^{2}-1$$

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

$$\text{true}\text{ and }0=2\times \left(\frac{\sqrt{2}}{2}\right)^{2}-1$$

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

$$\text{true}\text{ and }0=2\times \frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-1$$

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

$$\text{true}\text{ and }0=\frac{2\left(\sqrt{2}\right)^{2}}{2^{2}}-1$$

Cancel out 2 in both numerator and denominator.

$$\text{true}\text{ and }0=\frac{\left(\sqrt{2}\right)^{2}}{2}-1$$

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

$$\text{true}\text{ and }0=\frac{\left(\sqrt{2}\right)^{2}}{2}-\frac{2}{2}$$

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

$$\text{true}\text{ and }0=\frac{\left(\sqrt{2}\right)^{2}-2}{2}$$

The square of \sqrt{2} is 2.

$$\text{true}\text{ and }0=\frac{2-2}{2}$$

Subtract 2 from 2 to get 0.

$$\text{true}\text{ and }0=\frac{0}{2}$$

Zero divided by any non-zero number gives zero.

$$\text{true}\text{ and }0=0$$

Compare 0 and 0.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$