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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?

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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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  • $\begingroup$ Wow... I really need to improve if someone can look at this and answer it in a minute of looking at it. Thanks, this explains it perfectly. $\endgroup$ – Klik Sep 27 '14 at 16:44
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We can write $$A\sin(\theta)+B\cos(\theta)$$ in the form $C \sin(\theta+\phi)$ for some $\phi$ and $C$.

i.e. $$A\sin(\theta)+B\cos(\theta) \equiv C\sin(\theta+\phi).$$

Let's expand the RHS using the addition identity for sine.

$$A\sin(\theta)+B\cos(\theta) \equiv C\underbrace{[\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)]}_{\equiv \ \sin(\theta+\phi)}.$$

Simplifying, we have $$\color{red}A\sin(\theta)+\color{green}B\cos(\theta) \equiv \color{red}{C\cos(\phi)} \sin(\theta)+\color{green}{C\sin(\phi)}\sin(\theta).$$

This is an identity, so it's true for all (permitted) values of $\theta$.

Comparing coefficients of $\sin(\theta)$ we have $$A=C\cos(\phi) \tag{1}.$$

Comparing coefficients of $\cos(\theta)$ we have $$B=C\sin(\phi) \tag{2}.$$

If we do $\frac{(2)}{(1)},$ we have $$\frac{C\sin(\phi)}{C\cos(\phi)}=\Large \boxed{\tan(\phi)=\frac{\beta}{\alpha}}.$$

If we do $(1)^2+(2)^2,$ we get $$[C\cos(\phi)]^2+[C \sin(\phi)]^2 =C^2[\cos^2(\phi)+\sin^2(\phi)=C^2(1)=C^2 \quad.$$

Hence we have $$\Large \boxed{A^2+B^2=C^2} \quad.$$

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Hint : Just divide and multiply LHS by C and consider $\frac AC =\cos \phi$ and similarly $ \frac BC =\sin \phi$ and then try to simplify.

To get the proof consider

\begin{align} \sin (a-b) &= \cos (\pi/2-(a-b)) \\ &=\cos ((\pi/2-a)+b) \end{align} following on from this, one can then apply the appropriate formula for $\cos(a-b)$.

To find the proof of even this try searching on Google or follow this Wikipedia link. http://en.m.wikipedia.org/wiki/Proofs_of_trigonometric_identities

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