Calculus – Why A*sin(k(x+c)) Equals a*sin(kx) + b*cos(kx) Implies A=sqrt(a^2+b^2) and tan(c)=-b/a

calculustrigonometry

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 $\tan{kc}=-b/a$

Best Answer

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.

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