Learn How to Express cos(2x) as sin(x) Using Complex Exponential Series

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To express cos(2x) in terms of sin(x) using the complex exponential series, one can utilize the relationships e^{iθ} = cos(θ) + i sin(θ) and its derivatives. The relevant equations show that e^{iθ} + e^{-iθ} equals 2cos(θ) and e^{iθ} - e^{-iθ} equals 2i sin(θ). The complex exponential series is essentially the Taylor series for the exponential function with a complex variable. By substituting 2x into these equations, cos(2x) can be expressed in terms of sin(x). Understanding these relationships is key to solving the problem effectively.
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This is an easy question but can some else show/tell me how to do it:
"use the complex exponential series to express cos(2x) in terms of sin(x)"
I also don't quite understand the 'complex exponential series'. :redface:
 
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I mean can someone* show/tell me...
 
e^{i\theta} = \cos{\theta} + i\sin{\theta}

e^{i\theta} + e^{-i\theta} = 2\cos{\theta}

e^{i\theta} - e^{-i\theta} = 2i\sin{\theta}

cookiemonster
 
complex exponential series is just the taylor series for exp except with a complex variable instead of a real one
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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