Is it possible to set $c_1= c_1(a)$ and $c_2=c_2(a)$ such that for any $a>0$, for any two angles $\theta_i, \theta_j \in [0, 2\pi]$, we have that: $$ \log( 1 + a \cos(\theta_i - \theta_j) ) = c_1 \cos \theta_i \cos \theta_j + c_2 \sin \theta_i \sin \theta_j ? $$
Re-definition of scalar product $x \cdot y$ as $\log( 1 + a \, \, x \cdot y)$
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real-analysis
linear-algebra
functions
trigonometry
real-numbers
1 Answers
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No. Suppose the formula were true. Take $\theta_j=0$. Then $$ \log(1+a\cos\theta)=c_1(a)\cos\theta\quad\forall\theta\in[0,2\,\pi], $$ which is clearly not true.