# Using Euler's formula to prove trig identities using "sum to product" technique

• MiddleEast
In summary, the conversation discusses proving trigonometric identities using the Euler formula and the struggle to memorize sum to product identities. The author provides a hint for deriving these identities using the Euler formula, and suggests using the natural log as a tool for converting between sums and products.
MiddleEast
Homework Statement
Derive sum to product formulas using Euler formula
Relevant Equations
Euler formula e^ix = cos x + i sin x
Hello,
This is actually not homework.
I was google searching for "proving trig identities from geometric point of view), found one of the result which proves trig identities using Euler formula. I really liked it. Easier, quicker & simple.
But when the author speak about sum to product formulas, he gave unclear hint and did not do the derivation. Please see attached.
Can some1 explain how to do it using Euler? am really struggling with memorizing sum to product identities.
I still can derive them easily from angles sum & difference sin(x+y) ... etc, But I want to know how to do it using Euler formula.

Thanks

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What is meant is, for example, $$\begin{split} \cos a \cos b &= \frac{2\cos a \cos b}{2} + \frac{\sin a \sin b - \sin a \sin b}{2} \\ &= \frac{ \cos a \cos b + \sin a \sin b}{2} + \frac{\cos a \cos b - \sin a \sin b}{2} \\ &= \frac{\cos (a - b)}{2} + \frac{\cos (a + b)}{2} \end{split}.$$ But it is easier to start from $$\begin{split} \cos (a + b) &= \cos a \cos b - \sin a \sin b \\ \cos (a - b) &= \cos a \cos b + \sin a \sin b\end{split}$$ and add the two to get an expression for $\cos a \cos b$ or subtract the first from the second to get an expression for $\sin a \sin b$.

Thank you for your reply. Appreciated but it is algebraic. I need the one which uses euler formula.

Let $$e^{i(x+y)}=e^{ix}e^{iy}=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))$$ Likewise for $e^{i(x-y)}$. Then add and subtract.

PhDeezNutz and PeroK
As a general comment, the natural log takes you between sums and products, in that ln(ab)=ln(a) +ln(b)

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## What is Euler's formula and how is it related to trigonometric identities?

Euler's formula states that $$e^{ix} = \cos(x) + i\sin(x)$$. This relationship bridges complex exponential functions and trigonometric functions, providing a powerful tool for deriving and proving trigonometric identities.

## How can Euler's formula be used to derive the sum-to-product identities?

Euler's formula can be used to express trigonometric functions in terms of exponential functions. By manipulating these exponential expressions, we can derive the sum-to-product identities. For example, the sum of cosines can be written using Euler's formula and then simplified to show the sum-to-product form.

## What are the sum-to-product identities in trigonometry?

The sum-to-product identities are formulas that express sums or differences of trigonometric functions as products. They include: $$\cos(A) + \cos(B) = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$, $$\cos(A) - \cos(B) = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$, $$\sin(A) + \sin(B) = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$, $$\sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$.

## Can you provide a step-by-step example of proving a sum-to-product identity using Euler's formula?

Sure! Let's prove $$\cos(A) + \cos(B) = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$:1. Using Euler's formula: $$\cos(A) = \frac{e^{iA} + e^{-iA}}{2}$$ and $$\cos(B) = \frac{e^{iB} + e^{-iB}}{2}$$.2. Add these: $$\cos(A) + \cos(B) = \frac{e^{iA} + e^{-iA}}{2} + \frac{e^{iB} + e^{-iB}}{2}$$.3. Combine terms: \(\cos(A) + \cos(B) = \frac{e^{iA} + e^{iB} + e^{-iA} + e^{-iB}}{2

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