I would like to prove the following $$\cos(a+b)=\cos a\cdot\cos b - \sin a\cdot\sin b.$$
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1You can try to use $$\cos x=\frac{e^{ix}+e^{-ix}}2$$ and evaluate it in $x=a+b$. – Jul 24 '16 at 10:08
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3Yes we can ;o) But the way to do it depends on your definition of trigonometric functions – Bernard Jul 24 '16 at 10:09
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1Geometric proofs here. – Jul 24 '16 at 10:11
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Euler's formula states that $$e^{ix} = \cos(x)+i\sin(x)$$ Then $$e^{(a+b)i} = \cos(a+b) + i\sin(a+b)$$ but the above is also equal to $$e^{ai}e^{bi}=(\cos(a)+i\sin(a))(\cos(b)+i\sin(b)) =$$$$ (\cos(a)\cos(b)-\sin(a)\sin(b)) + i(\cos(a)\sin(b)+\sin(a)\cos(b))$$ And so the corresponding components are equal.
florence
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