I have come across this trig identity and I want to understand how it was derived. I have never seen it before, nor have I seen it in any of the online resources including the many trig identity cheat sheets that can be found on the internet.
$A\cdot\sin(\theta) + B\cdot\cos(\theta) = C\cdot\sin(\theta + \Phi)$
Where $C = \pm \sqrt{A^2+B^2}$
$\Phi = \arctan(\frac{B}{A})$
I can see that Pythagorean theorem is somehow used here because of the C equivalency, but I do not understand how the equation was derived.
I tried applying the sum of two angles identity of sine i.e. $\sin(a \pm b) = \sin(a)\cdot\cos(b) + \cos(a)\cdot\sin(b)$
But I am unsure what the next step is, in order to properly understand this identity.
Where does it come from? Is it a normal identity that mathematicians should have memorized?
Answer
The trigonometric angle identity $$\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$$ is exactly what you need. Note that $A^2 + B^2 = C^2$, or $$(A/C)^2 + (B/C)^2 = 1.$$ Thus there exists an angle $\phi$ such that $\cos\phi = A/C$ and $\sin\phi = B/C$. Then we immediately obtain from the aforementioned identity $$\sin(\theta+\phi) = \frac{A}{C}\sin \theta + \frac{B}{C}\cos\theta,$$ with the choice $\alpha = \theta$; after which multiplying both sides by $C$ gives the claimed result.
Note that $\phi = \tan^{-1} \frac{B}{A}$, not $\tan^{-1} \frac{B}{C}$.
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