0
$\begingroup$

This question already has an answer here:

I am lacking in understanding in the cosine and sine angle addition formulas. I have seen several questions similar to this but I have not seen an answer that explains how this conclusion can be derived. Geometric proofs offer little intuition and the rotation matrix is derived from the cosine and sin angle addition formulas. I haven't been able to derive this from calculus either and am already struggling on figuring out the sine and cosine derivatives in a way that makes sense instead of blindly accepting it. I would appreciate it if someone could answer why the cosine and sine angle addition formulas are as they are in an intuitive manner.

Thanks, Jackson

$\endgroup$

marked as duplicate by MCT, dustin, user147263, Jonas Meyer, Blue Mar 18 '15 at 1:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Some of the answers here may be of help (I'm marking it as a duplicate for now, but you may make your question more specific if you wish to make it not a duplicate) $\endgroup$ – MCT Mar 18 '15 at 1:14
  • $\begingroup$ It's a similar question but it's asking for something other than a geometric proof or just plain algebra. $\endgroup$ – Jackson H Mar 19 '15 at 0:14
0
$\begingroup$

You can. I approach trig via the unit circle. Here is the fundamental theorem of trig.

Theorem. Let $\theta\in\mathbb{R}$, $(x,y)\in\mathbb{R}^2$. Then, if you rotate the vector $xi + yj$ through angle $\theta$, the coordinates of the "head" of this vector will be $$(x\cos(\theta) - y \sin(\theta)), x\sin(\theta) + y\cos(\theta)).$$

To prove this, you should project the rotated vector onto $(x,y)$ and $(y, -x)$. You can do this via some simple triangle similarity arguments.

Once you have this, the addition formulae for sin and cos fall right out. See if you can think this through.

$\endgroup$
0
$\begingroup$

For example you want to prove:$$\cos (\alpha+\beta) = \cos (\alpha)\cos (\beta) - \sin (\alpha)\sin (\beta)$$.

You can use the identities: $$e^{i(\alpha+\beta)} = e^{i\alpha}\cdot e^{i\beta}$$,and

$$e^{i\alpha} = \cos (\alpha) + i\sin(\alpha)$$

$\endgroup$
  • $\begingroup$ See my comment in the answer by Qudit. $\endgroup$ – fredgoodman Feb 21 '18 at 15:19
0
$\begingroup$

One of the cleanest derivations is based on Euler's formula which tells us that $e^{i \theta} = \cos \theta + i \sin \theta$. The derivation works by noting that

\begin{align} \cos(x + y) + i \sin(x + y) & = e^{i (x + y)} \\ {} & = e^{ix} e^{iy} \\ {} & = (\cos x + i \sin x)(\cos y + i \sin y) \\ \end{align}

and expanding the product on the last line. I'll leave the details for you to work out.

$\endgroup$
  • $\begingroup$ It is difficult not to get involved in circular reasoning here. Where do you get Euler's formula? Perhaps by using power series. For which you certainly will need to at least know the derivatives of cos and sin. For which you will need the angle addition formulas. Or possibly you define $e^{i \theta}$ by Euler's formula, in which case the properties of the complex exponential follow from the angle addition formulas. Your suggestion is a good way to remember the angle addition formulas, however. $\endgroup$ – fredgoodman Feb 21 '18 at 15:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.