What is a nice proof of $$\cos(A) + \cos( B)= 2\cos\Big(\frac{A+B}{2}\Big)\cos\Big(\frac{A+B}{2}\Big)$$? I can prove it starting with the RHS but i want to be able to quickly prove it starting on the LHS as I won't have access to a formula book when I need it.
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I'm assuming you are allowed to use the cosine-of-a-sum formula? – Sinister Cutlass Dec 02 '15 at 19:19
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Yes i dont think i could forget that one if i wanted to – Robert S Dec 02 '15 at 19:20
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1You may also find the half-angle formula $\cos^{2}(\frac{x}{2})=\frac{1+\cos(x)}{2}$ useful. – Sinister Cutlass Dec 02 '15 at 19:21
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Maybe write $\cos(A)=\cos(A+B-B)$ and $\cos(B)=\cos(A+B-A)$ and unpack it from there? – Sinister Cutlass Dec 02 '15 at 19:23
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Do you know the identity $e^{ix}=\cos(x)+i\sin(x)$? – Surb Dec 02 '15 at 19:26
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Yes that's the reason why im looking for this proof. Need to add roots of unity in exponential form – Robert S Dec 02 '15 at 19:27
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yes that works, i changed (a+b-b) to (a+b/2 + a-b/2) though, works out a bit easier – Robert S Dec 02 '15 at 19:32
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With a change of variables, this answer has a picture proof. – Blue Dec 02 '15 at 23:40
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$$\cos(x+y)=\cos x\cos y- \sin x\sin y$$
$$\cos(x-y)=\cos x\cos y+\sin x\sin y$$
$$\cos(x+y)+\cos(x-y)=2\cos(x)\cos(y)$$
Now let $x+y=A$, $x-y=B$ and you'll get $$\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\left(\frac{A-B}{2}\right)$$
user236182
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Here is a fast and efficient way forward. We use Euler's Formula to write
$$\cos A+\cos B=\text{Re}\left(e^{iA}+e^{iB}\right) \tag 1$$
Now, simply factor out $e^{i(A+B)/2}$ under the "Real Part" operator in $(1)$ to obtain
$$\begin{align}\text{Re}\left(e^{iA}+e^{iB}\right) &=\text{Re}\left(e^{i(A+B)/2}\left(e^{i(A-B)/2}+e^{-i(A-B)/2}\right)\right) \\\\ &=\text{Re}\left(2\cos\left(\frac{A-B}{2}\right)e^{i(A+B)/2}\right) \\\\ &=2\cos\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right) \end{align}$$
And we are done!
Mark Viola
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This is a great solution. It took me a very long time to realize how to do this with Euler's formula. I kept trying to solve this while fully expanding each cosine into the real representation. Leaving it in the complex plane and taking the real part later is a great solution. +1 – akozi Nov 04 '20 at 18:34
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HINT: start from $$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$ $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$
Dr. Sonnhard Graubner
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Sorry i just realised my mistake there, made a typo in the question. I want a proof from the LHS – Robert S Dec 02 '15 at 19:22
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From LHS to RHS:
$$ \cos B + \cos A \\ \cos \frac {2 B} 2 + \cos \frac {2 A} 2 \\ \cos\left(\frac {A + B - A + B} 2 \right) + \cos\left(\frac {A + B + A - B} 2 \right) \\ \cos\left(\frac {A + B} 2 - \frac {A - B} 2 \right) + \cos\left(\frac {A + B} 2 + \dfrac {A - B} 2 \right) \\ \left[\cos\left(\frac{A + B} 2\right)\cos\left(\frac{A - B} 2 \right) + \sin\left(\frac{A + B} 2 \right)\sin\left(\frac{A - B} 2 \right)\right] + \left[\cos\left(\frac{A + B} 2 \right)\cos\left(\frac{A - B} 2 \right) - \sin\left(\frac{A + B} 2 \right)\sin\left(\frac{A - B} 2 \right)\right]\\ 2 \cos\left(\frac {A + B} 2 \right) \cos\left(\frac {A - B} 2 \right) \\ $$
cr3
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