I just had this thought. Can we find an identity that separates the product x.y in trig functions, i.e Cos(x.y)=f(x).g(y) or Sin(x.y) = f(x).g(y)

Question

I just had this thought. Can we find an identity that separates the product x.y in the trigonometric functions, i.e Cos(x.y)=f(x).g(y) or Sin(x.y) = f(x).g(y)... Where f(x) and g(y) are other trig functions,. Either as products or sums..?

Answer 1

The criteria for $\,A(x,y):=f(x)g(y)\,$ is that $$A(x_1,y_1)A(x_2,y_2) =f(x_1)g(y_1)f(x_2)g(y_2) = A(x_1,y_2)A(x_2,y_1) \tag{1} $$ always holds.

In the case of the $\,\cos\,$ function, suppose there exist functions $\,f,g\,$ such that $$\cos(x\,y)=f(x)g(y). \tag{2} $$ Then the equation $$ \cos(x_1y_1)\cos(x_2y_2) = \cos(x_1y_2)\cos(x_2y_1) \tag{3} $$ must be true. Specializing to the case $\,x_2=y_2=0\,$ reduces this equation to $\, \cos(x_1y_1) = 1 \,$ since $\,\cos(0)=1.\,$ However, this is not true in general and thus, there does not exist such functions $\,f,g.\,$

An similar result holds for the $\,\sin\,$ function. Specializing to the case $\,x_2=\pi,y_2=1\,$ reduces equation $(3)$ to $\,0=\sin(x_1)\sin(\pi y_1)\,$ since $\,\sin(\pi)=0.\,$ However, this is not true in general and thus, there does not exist such functions $\,f,g.\,$

Answer 2

In fact, we can go further; there are no (standard) trigonometric functions $f,g$ and no function $h$ with $\cos(xy) = h(f(x), g(y))$. Suppose there were; then we would have $1 = \cos(0)$ $= \cos((\frac14)\cdot 0)$ $= h(f(\frac14), g(0))$ $= h(f(\frac14), g(2\pi))$ $= \cos((\frac14)\cdot (2\pi))$ $= \cos(\frac\pi2) = 0$, where the middle equality is because $g(0) = g(2\pi)$ since $g$ has period $2\pi$.