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Use Euler's formula to derive the identity. (Note that if a, b, c, d are real numbers, a + bi = c + di means that a = c and b = d. Simplify your answer completely.) sin(2θ) = 2 sin(θ) cos(θ) Using Euler's formula, we have ei(2θ) = + i sin(2θ). On the other hand, ei(2θ) = (eiθ)2 = + i sin(θ) 2 = (cos2(θ) − sin2(θ)) + i sin(θ) . Equating Correct: Your answer is correct. parts, we find sin(2θ) = 2 sin(θ) cos(θ).

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Answer with Step-by-step explanation:

We have to prove that

[tex]sin 2\theta=2sin\theta cos\theta[/tex] by using Euler's formula

Euler's formula :[tex]e^{i\theta}=cos\theta+isin\theta[/tex]

[tex]e^{i(2\theta)}=(e^{i\theta})^2[/tex]

By using Euler's identity, we get

[tex]cos2\theta+isin2\theta=(cos\theta+isin\theta)^2[/tex]

[tex]cos2\theta+isin2\theta=(cos^2\theta-sin^2\theta+2isin\theta cos\theta)[/tex]

[tex](a+b)^2=a^2+b^2+2ab, i^2=-1[/tex]

[tex]cos2\theta+isin2\theta=cos2\theta+i(2sin\theta cos\theta)[/tex]

[tex]cos2\theta=cos^2\theta-sin^2\theta[/tex]

Comparing imaginary part on both sides

Then, we get

[tex]sin2\theta=2sin\theta cos\theta[/tex]

Hence, proved.

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