$\cos^2 (x) + \sin^2 (x) = \frac{(e^{ix} + e^{-ix})^2}{4} + \sin^2 (x) = \frac{(e^{2ix} + e^{-2ix})}{4} + e^{2\ln(\sin(x))} + \frac{1}{2}$

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Algebra 2

Factor sums and differences of cubes

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\cos^2 (x) + \sin^2 (x) = \frac{(e^{ix} + e^{-ix})^2}{4} + \sin^2 (x) = \frac{(e^{2ix} + e^{-2ix})}{4} + e^{2\ln(\sin(x))} + \frac{1}{2}

Pythagorean Identity: We know from the Pythagorean identity that $\cos^2(x) + \sin^2(x) = 1$ .
Left Side Simplification: So, the left side of the equation simplifies to $1$ .
Right Side Simplification: Now, let's simplify the right side. Start with the first term: $(e^{i\times} + e^{-i\times})^2 / 4$ .
First Term Simplification: Using the identity $(a + b)^2 = a^2 + 2ab + b^2$ , we get $(e^{ix})^2 + 2e^{ix}e^{-ix} + (e^{-ix})^2 / 4$ .
Term Division: Simplify each term: $e^{2ix} + 2 + \frac{e^{-2ix}}{4}$ .
Combining Like Terms: Now, divide each term by $4$ : $\frac{e^{2ix} + 2 + e^{-2ix}}{4} = \frac{e^{2ix}}{4} + \frac{1}{2} + \frac{e^{-2ix}}{4}$ .
Third Term Simplification: The second term on the right side is already simplified: $(e^{2ix} + e^{-2ix})/4$ .
Adding Terms: Combine the like terms from the first and second terms on the right side: $\frac{e^{2ix}}{4} + \frac{e^{-2ix}}{4} + \frac{e^{2ix}}{4} + \frac{e^{-2ix}}{4} + \frac{1}{2}$ .
Substitution: We get $\frac{2e^{2ix}}{4} + \frac{2e^{-2ix}}{4} + \frac{1}{2}$ .
Constant Combination: Simplify the coefficients: $e^{2ix}/2 + e^{-2ix}/2 + 1/2$ .
Isolating $\cos^2(x)$ : Now, let's look at the third term on the right side: $e^{2\ln(\sin(x))}$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .Combine the constants: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \frac{1}{2} + \frac{1}{2} - \cos^2(x)$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .Combine the constants: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \frac{1}{2} + \frac{1}{2} - \cos^2(x)$ .We get $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .Combine the constants: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \frac{1}{2} + \frac{1}{2} - \cos^2(x)$ .We get $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ .Now, we have $e^{\ln(\sin^2(x))}$ $0$ on the left side and $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ on the right side.
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .Combine the constants: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \frac{1}{2} + \frac{1}{2} - \cos^2(x)$ .We get $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ .Now, we have $e^{\ln(\sin^2(x))}$ $0$ on the left side and $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ on the right side.Subtract $e^{\ln(\sin^2(x))}$ $0$ from both sides to isolate $e^{\ln(\sin^2(x))}$ $3$ : $e^{\ln(\sin^2(x))}$ $4$ .
Final Simplification: Using the property $e^{\ln(a)} = a$ , we get $e^{\ln(\sin^2(x))}$ .This simplifies to $\sin^2(x)$ .Add this to the previous terms: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \sin^2(x) + \frac{1}{2}$ .We know $\sin^2(x) = 1 - \cos^2(x)$ from the Pythagorean identity.Substitute $\sin^2(x)$ with $1 - \cos^2(x)$ : $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x) + \frac{1}{2}$ .Combine the constants: $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + \frac{1}{2} + \frac{1}{2} - \cos^2(x)$ .We get $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ .Now, we have $e^{\ln(\sin^2(x))}$ $0$ on the left side and $\frac{e^{2ix}}{2} + \frac{e^{-2ix}}{2} + 1 - \cos^2(x)$ on the right side.Subtract $e^{\ln(\sin^2(x))}$ $0$ from both sides to isolate $e^{\ln(\sin^2(x))}$ $3$ : $e^{\ln(\sin^2(x))}$ $4$ .This simplifies to $e^{\ln(\sin^2(x))}$ $5$ .