Transforming $\tan{\phi} = c_1 \tan{\phi_1}+ c_2 \tan{\phi_2}$ into the form $\phi = d_1 \phi_1 + d_2 \phi_2$

Question

my problem is that I am trying to transform this trigonometric expression: $$\tan{\phi} = c_1 \tan{\phi_1}+ c_2 \tan{\phi_2}$$

to the form: $$\phi = d_1 \phi_1 + d_2 \phi_2$$

Note: Here $c_1$ = $\frac{\beta_1}{\beta_1+\beta_2}$ and $c_2$ = $\frac{\beta_2}{\beta_1+\beta_2}$

My basic aim is to represent $c_1$ as $f_1(d_1,d_2, \phi_1, \phi_2, \phi)$ and $c_2$ as $f_2(d_1,d_2, \phi_1, \phi_2, \phi)$ by comparing the coefficients. But the final form is necessary meaning that the comparison of coefficients is done only after the given form $d_1 \phi_1 + d_2 \phi_2$ is attained on the RHS. I would be really grateful if someone could help me out with this. Also if someone could suggest someother method for this function estimation, that could also be of great help. Thanks in advance.

Answer 1

To simplify the notation, let's write $$ \tan z = a\tan x + b\tan y $$

Then, for $-\pi / 2 < x,y < \pi/2$ we want to express $z$ as $$ z(x,y) = \arctan (a\tan x + b\tan y) = f(x,y)x + g(x,y)y $$ (we omit to indicate $a$ and $b$ among the variables of the functions).

As a first step, we have that $$ \left\{ \matrix{ z(0,0) = 0 \hfill \cr z(x,0) = \arctan (a\tan x) = f(x,0)x \hfill \cr z(0,y) = \arctan (b\tan y) = g(0,y)y \hfill \cr} \right. $$ Expanding into series $$ \eqalign{ & f(x,0)x = z(x,0) = \arctan (a\tan x) = \cr & = ax - {a \over 3}\left( {a^{\,2} - 1} \right)x^{\,3} + O\left( {a^{\,5} x^{\,5} } \right)\quad \Rightarrow \cr & \Rightarrow \quad f(x,0) = a - {a \over 3}\left( {a^{\,2} - 1} \right)x^{\,2} + O\left( {a^{\,5} x^{\,4} } \right) \cr & \Rightarrow \quad g(0,y) = b - {b \over 3}\left( {b^{\,2} - 1} \right)y^{\,2} + O\left( {b^{\,5} y^{\,4} } \right) \cr} $$

That hints to the fact that we can write $$ \eqalign{ & z(x,y) = \arctan (a\tan x + b\tan y) = \cr & = ax + by + \left( {{{\arctan (a\tan x + b\tan y) - ax - by} \over {x + y}}} \right)\left( {x + y} \right) = \cr & = \left( {{{\arctan (a\tan x + b\tan y) + \left( {a - b} \right)y} \over {x + y}}} \right)x + \cr & + \left( {{{\arctan (a\tan x + b\tan y) + \left( {b - a} \right)x} \over {x + y}}} \right)y \cr} $$

Answer 2

If i understand right, you can write

$$\frac{\tan(\phi_1)}{c_1^2+c_2^2}+\frac{c_2}{c_1^2+c_2^2}\tan(\phi_2)$$