I don't understand this. These identities are given in the online notes for MIT's 18.01 calculus class. It's related to taking the sum of two trig functions and transforming them into a single trig function. I can use the formulas to do this, but I am having trouble finding anything on the internet to use a proof for my understanding.

Thanks for any responses!

EDITED (added after the first answer):

Sorry, the identity goes like this:

$A\sin{k(x-c)}=a\sin{kx}+b\cos{kx}$ ,

with the relationships

$a=A\cos{kc}$ , $b=-A\sin{kc}$ , $A=\sqrt{a^2+b^2}$ , and $\tan{kc}=-b/a$

  • 1
    $\begingroup$ This answer of mine walks through the process of creating a picture-proof of the identity (where $k=1$). $\endgroup$ – Blue Jul 4 '14 at 20:46
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    $\begingroup$ @Blue - you should copy your diagram and post it as an answer. $\endgroup$ – nbubis Jul 4 '14 at 20:47

You might want $\sin (kx+c)$ rather than $\sin k(x+c)$, and lose a minus sign on the $\tan$ expression, but the shape of the argument is as follows:

We start by noting that $A\sin (P+Q)=A\sin P\cos Q+A\cos P\sin Q$, for any values of $P,Q$ and we choose to use $P=kx$ and $Q=kc$

Now equating coefficients on the right-hand side we need $a=A\cos Q$ and $b=A\sin Q$ whence $a^2+b^2=A^2(\sin^2 Q+\cos^2 Q)=A^2$, and we also have $\tan Q=\cfrac ba$

This would give you what you wanted if you had $Q=c$ rather than $Q=kc$ and changed the sign - which is why I suggested checking the question carefully.


Here's my picture-proof of the identity, with $k=1$ and @MarkBennet's suggestion to remove the negative sign from the tangent:

enter image description here

$$p \sin(\bullet) + q \cos(\bullet) = r \sin(\bullet +\circ ), \quad\text{where}\quad r = \sqrt{p^2+q^2} \quad\text{and}\quad \tan(\circ) = \frac{q}{p}$$

I walk through the creation of the diagram in this answer.

  • $\begingroup$ FYI, I posted a version of this to trigonography.com. $\endgroup$ – Blue Dec 25 '15 at 19:08
  • $\begingroup$ Awesome visualization. I wish all math proofs can be as intuitive as this. $\endgroup$ – qed Jan 1 '17 at 13:46

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