Sin(a+b)/Cos(a+b) Derivations? (Straightforward)

[jegues]

I'm just curious about the,

sin(a\pm b), \quad cos(a\pm b)

identities, is there a simple way to derive them?

I have a test in a few days and I don't like memorizing too much.

Thanks again!

 

[Char. Limit]

Here are a couple of videos that may help:

http://www.khanacademy.org/video/proof--sin-a-b------cos-a--sin-b-----sin-a--cos-b?playlist=Trigonometry

http://www.khanacademy.org/video/proof--cos-a-b-----cos-a--cos-b---sin-a--sin-b?playlist=Trigonometry

 

[sponsoredwalk]

If you go to page 272 of this book, the trig chapter, there is a nice derivation requiring
very little memory or crazy drawings requiring superhuman memorization. I
can't really see the pictures on googlebooks so go to page 137 of this book to get a second perspective
(which is really the exact same one with visible drawings because Lang is known for
copying almost verbatim from one section of a book he wrote to another ).

Ps: That calc book is really recommended to cut down on a lot of memorization

edit: Man that derivation is so amazing! I haven't looked at that in so long & totally forgot it but
had been meaning to. Classic Lang, such a simple derivation of something complex
using a trivial idea like tilted coordinate axes & some elbow grease

 

[chiro]

I'm just curious about the,

sin(a\pm b), \quad cos(a\pm b)

identities, is there a simple way to derive them?

I have a test in a few days and I don't like memorizing too much.

Thanks again!

I don't know if this has been mentioned in the hyperlinks but you can use Euler's formula to derive trigonometric identities.

Basically eulers formula is e^(i x theta) = cos(theta) + i x sin(theta). Using the binomial theorem and eulers formula you can equate the cos(theta) and sin(theta) with another decomposition.

Its quite remarkable in my mind that this even works as good as it does and you need no geometry to do it.

 

[HallsofIvy]

A method I like is this- we know from Calculus that (sin(x))'= cos(x) and (cos(x))'= -sin(x). Differentiating again, (sin(x))''= (cos(x))'= -sin(x) and (cos(x))''= (-sin(x))'= - cos(x). That is, both y= sin(x) and y= cos(x) satisfy the differential equation y''= -y.

That is a linear second order homogeneous differential equation so any solution can be written as a linear combination of two independent solutions. sin(x) and cos(x) are independent because one is not just a multiple of the other. That is, any solution to the differential equation y''= -y can be written in the form y= Acos(t)+ B sin(t) for some numbers A and B. Further, it is easy to see that y(0)= Acos(0)+ Bsin(0)= A and y'(0)= -Asin(0)+ Bcos(0)= B. That is, the coefficients are just the function and its derivative evaluated at 0.

Now, look at y(x)= sin(x+ b). Then y'= cos(x+ b) and y''= -sin(x+ b)= -y. That is, sin(x+ b) also satisfies y''= -y. Further, y(0)= sin(b) and y'(0)= cos(b). We can write sin(x+ b)= sin(b)cos(x)+ cos(b)sin(x). Taking x= a, sin(a+ b)= sin(b)cos(a)+ cos(b)sin(a).

Similarly, let y(x)= cos(x+ b). Then y'= -sin(x+ b) and y''= - cos(x+ b)= y. Now y(0)= cos(b) and y'(0)= - sin(b) so we have
y(x)= cos(x+ b)= cos(b)cos(x)- sin(b)sin(x).

Taking x= a, cos(a+ b)= cos(b)cos(a)- sin(b)sin(a).

 

[sponsoredwalk]

This isn't an argument against a derivation involving derivatives but the derivations of
derivatives of sin & cos that I've seen all involve double angle formula's in the middle of
the derivation of the derivative in the first place so derivatives presuppose knowledge of
that which you're trying to discover.

As I said it's not an argument against it but if you're deriving things based on
as little knowledge as possible it's good to be aware. I see the creativity in deriving
things different ways & both the e^i & diff eq derivations are really cool, just thought
I'd mention that

 

[hunt_mat]

What about using:
<br /> e^{i\theta}=\cos\theta +i\sin\theta<br />
then use:
<br /> e^{i\theta}e^{i\phi}=e^{i(\theta +\phi )}<br />

 

[HallsofIvy]

This isn't an argument against a derivation involving derivatives but the derivations of
derivatives of sin & cos that I've seen all involve double angle formula's in the middle of
the derivation of the derivative in the first place so derivatives presuppose knowledge of
that which you're trying to discover.

As I said it's not an argument against it but if you're deriving things based on
as little knowledge as possible it's good to be aware. I see the creativity in deriving
things different ways & both the e^i & diff eq derivations are really cool, just thought
I'd mention that

Good point but one method of defining sine and cosine is

Define y(x)= sin(x) as the solution to the differential equation y''+ y= 0 satifying the initial conditions y(0)= 0, y'(0)= 1.

Define y(x)= cos(x) as the solution to the differential equation y''+ y= 0 satisfying the initial conditions y(0)= 1, y'(0)= 0.

Another way to define sine and cosine is
sin(x)= \displaytype \sum_{n= 0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}
and
cos(x)= \displaytype \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}
And the derivatives can be done by differentiating "term by term".

All properties of sine and cosine, including periodicity, can be proved from those definitions.

 

[snipez90]

Protip: memorize them. Sure if you've taken one of the suggestions here and can rederive them in a minute, more power to you. But that derivation costs a minute (possibly more) on the exam, when you could have simply wrote down the formula. Seeing the derivation is helpful, but it's silly to re-derive basic formulas every time you need them. Using the formula enough times on homework is the easiest way to remember the formula, but if you haven't, please memorize the formula.