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$a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,

where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.

I want to know how to get to this result. I'm able to derive $c$ by taking the derivative of the equation, then squaring both and adding them together, and back-substituting the cosine of a double angle.

But how does one get to the expression for $\tan(\phi)$?

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  • $\begingroup$ Do you know of dot products? I know of a beautiful way to incorporate this here. $\endgroup$ Jul 8, 2017 at 18:53
  • $\begingroup$ I'm familiar with the dot product of vectors, if this is what you mean. $\endgroup$
    – Petoetje59
    Jul 8, 2017 at 18:55

4 Answers 4

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Use the sin addition formula $\sin(\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$ \begin{eqnarray*} a \sin x + \underbrace{b \sin(x+\theta)}_{ b\sin x \cos \theta+b \cos x \sin \theta}= \underbrace{c \sin(x+ \phi)}_{b\sin x \cos \phi+b \cos x \sin \phi} \\ (a + b \cos \theta) \color{red}{\sin x} + b \sin \theta \color{blue}{\cos x} = c \cos \phi \color{red}{\sin x}+c \sin \phi \color{blue}{\cos x} \end{eqnarray*} Equate the coefficients of $ \sin x $ and $ \cos x $ \begin{eqnarray*} a + b \cos \theta = c \cos \phi \\ b \sin \theta =c \sin \phi \end{eqnarray*} Now square these equations and add to get the first equation you want. & take the ratio of these equations to get the second equation.

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  • $\begingroup$ Yep, that's it. Thanks ! $\endgroup$
    – Petoetje59
    Jul 8, 2017 at 19:09
  • $\begingroup$ +1. Very nice. You could/should comment about why equating the coefficients of $\sin x$ and $\cos x$ is valid (which, of course, it is). $\endgroup$
    – Blue
    Jul 8, 2017 at 19:40
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For fun, here's a trigonograph, which leads to a counterpart cosine identity:

enter image description here

$$\begin{align} a \sin x + b \sin(\theta+x) &= c\sin(\phi+x) \\ a \cos x + b \cos(\theta+x) &= c\cos(\phi+x) \end{align}$$ where $$c^2 = a^2 + b^2 - 2 a b \cos(180^\circ-\theta) = a^2 + b^2 + 2 a b \cos \theta \qquad\text{and}\qquad \tan\phi = \frac{b\sin\theta}{a+b\cos\theta}$$

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  • $\begingroup$ Nice geometric proof, but nothing beats Splutterwit's simplest solution. $\endgroup$
    – Petoetje59
    Jul 8, 2017 at 19:30
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We use $a \cdot b=|a||b|\cos(\text{angle between vectors})$.

Note,

$$\sin (x+\theta)=\sin x \cos \theta+\sin \theta \cos x$$

Hence the left hand side of your equation is,

$$a\sin x+b\sin x \cos \theta+b \cos x \sin \theta$$

$$=(a+b\cos \theta)\sin x+(b\sin \theta)\cos x$$

$$=\langle b\sin \theta,a+b\cos \theta \rangle \cdot \langle \cos x,\sin x \rangle$$

The angle between these two vectors is $\alpha-x$ or $x-\alpha$ where $\cot \alpha=\frac{b \sin \theta}{a+b\cos \theta}$, depending on the relative positioning of these vectors. In any case the cosine of the angle between these two vectors is $\cos(\alpha-x)=\cos(x-\alpha)$ do to the fact cosine is even.

Next note that $\sin (x+\frac{\pi}{2})=\cos(x)$ so that the cosines between the two angles is really:

$$\sin (x-\alpha+\frac{\pi}{2})$$

But:

$$\tan (\frac{\pi}{2}-\alpha)=\cot \alpha$$

So,

$$\tan (\phi)=\cot \alpha=\frac{b \sin \theta}{a+b\cos \theta}$$.

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  • $\begingroup$ I can follow your reasoning, but where is the $\phi$? Splutterwit's solution is clear to me. An alternative solution is always welcome. $\endgroup$
    – Petoetje59
    Jul 8, 2017 at 19:16
  • $\begingroup$ I'll edit @Petoetje59 $\endgroup$ Jul 8, 2017 at 19:17
  • $\begingroup$ To precise we must examine the cases $\alpha=0,\pi,2\pi$ separately. If $\alpha=0$ then $a+b\cos \theta=0$,etc $\endgroup$ Jul 8, 2017 at 20:12
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you will have $$\frac{a\sin(x)+b\sin(x+\theta)}{c}=\sin(x+\phi)$$ $$\arcsin\left(\frac{a\sin(x)+b\sin(x+\theta)}{c}\right)-x=\phi$$

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  • $\begingroup$ Obviously, but how does one obtain the much simpler expression for $\tan(\phi)$ i.e. without the dependence on $x$? I need this solution. $\endgroup$
    – Petoetje59
    Jul 8, 2017 at 18:50

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