# The Derivation of Several Triginometric Identities

1. May 20, 2015

### Drafter

I would like to know how can I derive the double, half, product to sum, and sum to product identities of trigonometry using simple algebraic means.

And which books (which I prefer) should I pickup at the library on this subject to actually learn these derivations? Or at the very least some videos?
I don't like reading text online, I find it hard to follow and very passive and unengaging in my opinion.

2. May 20, 2015

### Staff: Mentor

3. May 20, 2015

### Drafter

In the wiki section the sum identities did go through a nice proof, but i'm not too keen on deriving the sum in of itself, however I am interested on finding a sum to product proof though.
.
I just found it silly how my professor told me to memorize "this" and memorize "that".
"You're better off memorizing these formulas, it's really hard to derive them."

I think deriving them through logical progression would be, albeit a bit longer, but would of surely been easier to do; rather than relying on memory that could potentially blank out in the heat of an exam. It would make sense to test students on their knowledge of proofs and problem solving, rather than memorization and problem solving in my opinion.

Because of that, i'm itching to find a way to derive these half angle, double angle, sum to product, and product to sum identities. Because the proofs can't be all that bad I suppose.

Idk, I guess i'll check a library and see what they have in store.

4. May 20, 2015

### pasmith

There are two approaches. The first is geometric: draw the appropriate triangles and find lengths of sides.

The second is analytic: use the definitions of cosine and sine in terms of the exponential function and do some algebra.

5. May 20, 2015

### Svein

That's where complex analysis does such a great job. For example $e^{i\phi} = \cos (\phi) + i\cdot \sin(\phi)$. Therefore $e^{i2 \phi} = \cos (2\phi) + i\cdot \sin(2\phi)$, but also $e^{i2 \phi} = (\cos (\phi) + i\cdot \sin(\phi))^{2} = \cos^{2} (\phi)+2i\cos (\phi)\sin(\phi) -\sin^{2}(\phi)$. Now sort the real part and the imaginary part and get $\cos (2\phi) = \cos^{2} (\phi) -\sin^{2}(\phi)$ and $\sin(2\phi) = 2\sin(\phi)\cos(\phi)$. In the same way you can calculate cos(x+y) and sin(x+y). Half the angle: Put θ = φ/2, then φ=2⋅θ...

6. May 20, 2015

### Staff: Mentor

My preference was doing the proof via a diagram for a+b and a-b and then letting a=b to get the doubling identities. I felt using the Euler formula was using a higher level of math to do something that is more basic.

7. May 23, 2015

### HallsofIvy

Since Svein refers to writing sine and cosine in terms of the exponential, I feel free to refer to yet another, somewhat unusual, definition of sine and cosine:

We define y= sin(x) as the unique solution to the differential equation y''+ y= 0 with initial conditions y(0)= 1, y'(0)= 0 and define y= cos(x) as the unique solution to the same differential equation with initial conditions y(0)= 0, y'(0)= 1.

With those initial conditions, it is easy to show that they are "independent solutions" and so, because this is a second order linear differential equation, any solution can be written as a linear combination of those two functions. In fact, it is easy to see that the solution to the equation y''+ y= 0 with initial conditions y(0)= A, y'(0)= B is y(x)= A cos(x)+ B sin(x).

Now, let y= sin(x+ a) for constant a. Then y'= cos(x+a) and y''= -sin(x+ a) so that y''+ y= sin(x+ a)- sin(x+ a)= 0 while y(0)= sin(a) and y'(0)= cos(a). That is, sin(x+ a) is a solution to the initial value problem y''+ y= 0 with initial condition y(0)= sin(a), y'(0)= cos(a). So it follows that sin(x+ a)= sin(a)cos(x)+ cos(a)sin(x). Similarly, taking y= cos(x+ a), y'= -sin(x+ a) and y''= -cos(x+ a) while y(0)= cos(a) and y'(0)= -sin(a). It follows that cos(x+ a)= cos(a)cos(x)- sin(a)sin(x).

Taking, in both of those, x= b, we have sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b) and cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

From those "addition formulas" the others follow:
Letting a= b= x, sin(2x)= sin(x+ x)= sin(x)cos(x)+ cos(x)sin(x)= 2 sin(x)cos(x) and cos(2x)= cos(x)cos(x)- sin(x)sin(x)= cos^2(x)- sin^2(x).

Of course, sin^2(x)+ cos^2(x)= 1 so we can write sin^2(x)= 1- cos^2(x) so that we can write that last formula as cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1= cos(2x). Replacing "x" with "x/2", that becomes 2cos^2(x/2)- 1= cos(x) and, solving for cos(x/2), we have cos(x/2)= sqrt((1+ cos(x)/2). Similarly, cos^2(x)= 1- sin^2(x), so cos(2x)= cos^2(x)- sin^2(x)= 1- 2sin^2(x) so, replacing x with x/2 and solving for sin(x/2) we have sin(x/2)= sqrt((1- cos(x))/2).

8. May 31, 2015

### Drafter

I do appreciate the way you the identities algebraically here........
Everything else seemed to be a little over my head sadly, since I have just started walking into the doorsteps of the calculus.

I could see that the derivative and second derivative of those trig functions were very nice, sadly I have only just started with the dervative of very simple, plane jane functions, in fact, the method of deratives I use aren't even formalised yet, it's a very redmunetary form called method of incrememnts, that I learned in book i've been reading.

But nevertheless, the proof I qouted here is golden. I thank you.

I bought a trig book too that pretty much stated the same thing you just said here for the double and half angle identities, nice.