Trigonometric Identities Like $A \sin(x) + B \cos(y) = \cdots$

Question

Are there any identities for trigonometric equations of the form:

$$A\sin(x) + B\sin(y) = \cdots$$ $$A\sin(x) + B\cos(y) = \cdots$$ $$A\cos(x) + B\cos(y) = \cdots$$

I can't find any mention of them anywhere, maybe there is a good reason why there aren't identities for these? Thanks!

Answer 1

Since

$$ A \cos(a+b) = A \cos(a) \cos(b) - A \sin(a) \sin(b) \ \ \ \ \ \ (1) \\ B \cos(a-b) = B \cos(a) \cos(b) + B \sin(a) \sin(b) \ \ \ \ \ \ (2) $$

(1) + (2) gives

$$ A \cos(x) + B \cos(y) = (A+B) \cos(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + (B-A) \sin(\frac{x+y}{2}) \sin(\frac{x-y}{2}) $$

where

$$ x = a + b \\ y = a - b $$

substitute

$$ Q = (A+B) \cos(\frac{x-y}{2}) \\ R = (B-A) \sin(\frac{x-y}{2}) \\ P = \frac{x+y}{2} $$

then

$$ A \cos(x) + B \cos(y) = Q \cos P + R \sin P = \sqrt{Q^2+R^2} \cos(P-\phi) $$

where

$$ \sin \phi = \frac{R}{\sqrt{Q^2+R^2}} \\ \cos \phi = \frac{Q}{\sqrt{Q^2+R^2}} $$

Answer 2

there are no general formula for these expressions.but may exist when $A$ and $B$ are interrelated .

For example consider triangle $ABC$ where $a,b,\text{ and }c $ are the sides of the triangle and $A,B,\text{ and }C$ are the respective angles opposite to $a,b,\text{ and }c $ then $$c = a\cos B + b\cos A $$ here this is because $a,b ,A\text{ and }B$ are interrelated by laws of triangle.

therefore random values of the angles and the coefficients will not satisfy to form general formula.

Answer 3

Since the proof was asked for Asin(x)+Bsin(y), based on the prior answer, I expand this time with a Python example for some verification:

$$ A \sin(a+b) = A \sin(a) \cos(b) + A \cos(a) \sin(b) \ \ \ \ \ \ (1) \\ B \sin(a-b) = B \sin(a) \cos(b) - B \cos(a) \sin(b) \ \ \ \ \ \ (2) $$

(1) + (2) gives

$$ A \sin(x) + B \sin(y) = (A+B) \sin(\frac{x+y}{2}) \cos(\frac{x-y}{2}) + (A-B) \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2}) $$

where

$$ x = a + b \\ y = a - b $$

substitute

$$ Q = (A+B) \sin(\frac{x+y}{2}) \\ R = (A-B) \cos(\frac{x+y}{2}) \\ P = \frac{x-y}{2} $$

then

$$ A \sin(x) + B \sin(y) = Q \cos P + R \sin P = \sqrt{Q^2+R^2} \sin(P+\phi) $$

where

$$ \tan \phi = \frac{Q}{R} \\ $$

Proof in Python:

import numpy as np
A, B, x, y = 3, 4, 5, 6
Q, R, P = (A+B)*np.sin((x+y)/2), (A-B)*np.cos((x+y)/2), (x-y)/2
print(A*np.sin(x)+B*np.sin(y), np.sqrt(Q*Q+R*R)*np.sin(P+np.arctan2(Q, R)))

Result:

-3.9944348167851187 -3.9944348167851187

Answer 4

$A \, \cos(x) + B \, \cos(y)= C \, \cos(z)$, where, $$ C = \sqrt{(A \, \cos(x) + B \, \cos(y))^2 + (A \, \sin(x) + B \, \sin(y))^2}, $$ and $$ z = \tan^{-1}\left(\frac{A \, \sin(x) + B \, \sin(y)}{A \, \cos(x) + B \, \cos(y)}\right). $$