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I'm having a lot of trouble in the following task: consider the real numbers $A,B,C,a,b_1,b_2,b_3 \in \mathbb{R}$ such that the following trigonometric equation holds for every $x \in \mathbb{R}$:

$A\cos(ax+b_1)+B\cos(ax+b_2)=C\cos(ax+b_3)$

with $A,B,C,a>0$. Prove that $A+B=C$.

$ $

My incomplete attempt:

Let's differentiate both sides wrt x. Then we have:

$A\sin(ax+b_1)+B\sin(ax+b_2)=C\sin(ax+b_3)$

Taking the square of both sides in both the equations and then adding them together we obtain:

$\cos(b_1-b_2)=\frac{A_3^2-A_1^2-A_2^2}{2A_1A_2}$. Now if $b_1=b_2$ we have that $\cos(b_1-b_2)=1$ and then $A_1+A_2=A_3$. However, here I'm taking $b_1=b_2$ without any reason. Every effort I make to show that $b_1=b_2$ take me, after a lot of calculation, to an identity. Thank you for any help or hint!

Leonardo
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5 Answers5

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Prove this holds for all $x$: $$ \cos(x + \pi/3) + \cos(x - \pi/3) = \cos(x) $$

GEdgar
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  • So this means that if we take $A=1,B=1,C=1,a=1,b_1=\frac{\pi}{3},b_2=-\frac{\pi}{3},b_3=0$, then my equation holds for every x but we have that $A+B \neq C$, right? – Leonardo Sep 20 '21 at 16:14
  • @Leonardo Re GEdgar's answer, which is based on the idea that $\cos(\pi/3) = \cos(-\pi/3) = (1/2), (1/2) + (1/2) = 1$, can be expanded to consideration of $\cos(x + \theta), \cos(x - \theta)$. Simply choose $A = B = \frac{1}{2\cos(\theta)}.$ Note that $\sin(x)\sin(\theta) + \sin(x)\sin(-\theta) = 0.$ – user2661923 Sep 20 '21 at 16:29
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Rewrite the constraint as

$\{\cos(ax) ~~~\times~~~ [A\cos(b_1) + B\cos(b_2) - C\cos(b_3)]\}$
$-~~~ \{\sin(ax) ~~~\times~~~ [A\sin(b_1) + B\sin(b_2) - C\sin(b_3)]\} ~~~=~~~ 0.$

If $a = 0$, then the above constraint reduces to $A\cos(b_1) + B\cos(b_2) - C\cos(b_3) = 0.$

Without loss of generality, $a \neq 0.$

In effect, you have two constants $E,F$ such that for all values of $(x)$, the following equation must hold.

$$E\cos(ax) - F\sin(ax) = 0.\tag1 $$

Clearly, if $E = 0$, then $F$ must also $= 0$, and vice versa. Suppose that $E \neq 0 \neq F.$

Then equation (1) above can not hold for all values of $(x)$, because the function $f(x) = \tan(x)$ is not a constant function.

Therefore, the question reduces to finding constants $A, B, C, b_1, b_2, b_3$ such that

$$[A\cos(b_1) + B\cos(b_2) - C\cos(b_3)] = 0 = A\sin(b_1) + B\sin(b_2) - C\sin(b_3).]\tag2$$

One obvious solution is
$0= b_1 = b_2 = b_3$ and
$A + B - C = 0.$

Personally, I am not sure how to exhaustively find all values $A,B,C,b_1, b_2, b_3$ such that (2) above is satisfied.

user2661923
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    $C \cos b_3=A \cos b_1+B\cos b_2$, $C \sin b_3=A \sin b_1+B\sin b_2$

    The system has the solution $(C\geq 0, b_3)$ at any $A > 0, B > 0, b_1, b_2$, so your solution is not unique.

     – Ivan Kaznacheyeu Sep 20 '21 at 16:13
  • @IvanKaznacheyeu Agreed - hence : "one obvious solution...". Also, the answer of GEdgar, where (in effect) $b_1 = \pi/3, b_2 = -\pi/3, b_3 = 0$ can be used to generate a solution. As I indicated in my answer, I don't know how to generate all solutions. – user2661923 Sep 20 '21 at 16:16
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If, with $y=ax$, $$A\cos(y+b_1)+B\cos(y+b_2)-C\cos(y+b_3)=0 \qquad \forall x$$ it must hold for small values of $y$.

Then, using the Taylor series around $y=0$, we should have $$(A \cos (b_1)+B \cos (b_2)-C \cos (b_3))- (A \sin (b_1)+B \sin (b_2)-C \sin (b_3))\,y+O\left(y^2\right)=0$$

Solve the two linear equations for $A$ and $B$. This gives after trigonometric simplification $$A=-C \sin (b_2-b_3) \csc (b_1-b_2)$$ $$B=C \sin (b_1-b_3) \csc (b_1-b_2)$$ $$A+B=C \csc (b_1-b_2)\,(\sin (b_1-b_3)-\sin (b_2-b_3))$$ $$A+B= C \cos \left(\frac{b_1+b_2}{2} - b_3\right) \sec \left(\frac{b_1-b_2}{2} \right)$$

So, the general condition to have $A+B=C$ is given by $$\cos \left(\frac{b_1+b_2}{2} - b_3\right)= \cos \left(\frac{b_1-b_2}{2}\right)$$ that is to say $$ \sin \left(\frac{b_1-b_3}{2} \right) \sin \left(\frac{b_2-b_3}{2}\right)=0$$

Claude Leibovici
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We indicate with $u=cos(a x+b_{1})$, $v=cos(a x+b_{2})$ and $w=cos(a x+b_{3})$.

The equations become:

$u A+v B=w C$,

$A+B=C$.

We will consider the system of two equations in the unknowns $A$ and $B$. Solving it we get:

$A=C \frac{w-v}{u-v}$,

$B=C \frac{u-w}{u-v}$.

Verification: $A+B= C \frac{w-v}{u-v}+ C \frac{u-w}{u-v}=C (\frac{w-v}{u-v}+\frac{u-w}{u-v})=C \frac{w-v+u-w}{u-v}=C$

Finally, by giving $u$, $v$ and $w$ their values, we get:

$A=\frac{cos(ax+b_{3})-cos(ax+b_{2})}{cos(ax+b_{1})-cos(ax+b_{2})}$,

$B=\frac{cos(ax+b_{1})-cos(ax+b_{3})}{cos(ax+b_{1})-cos(ax+b_{2})}$.

giuseppe mancò
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$$A\cos(ax+b_1)+B\cos(ax+b_2)=C\cos(ax+b_3) =C\cos(ax+b_3)~~~~(1)$$ Note that if $$P \sin uA+ Q \cos uA=0, \forall x \in R \implies P=0,Q=0$$ Because $\sin uA, cos uA$ are linearly independent. $$A \cos ax \cos b_1-A \sin ax \sin b_1+ B\cos ax \cos b_2- B\sin ax \sin b_2= C\cos ax \cos b_3- C \sin ax \sin b_3$$ $$\implies \cos ax (A \cos b_1+B \cos b_2-C \cos b_3)+ \sin ax (-A \sin b_1-B \sin b_2+ C\sin b_2)=0$$ $$\implies (A \cos b_1+B \cos b_2-C \cos b_3)=0=(-A \sin b_1-B \sin b_2+ C\sin b_2)$$

Z Ahmed
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