Please provide proof for the main trig functions

Hi MHF. If anyone has time, I would love for concrete proof for the following trig functions:

$\displaystyle sin(u+v)=sin(u)\cdot cos(v)+cos(u)\cdot sin(v)\\sin(u-v)=sin(u)\cdot cos(v)-cos(u)\cdot sin(v)\\cos(u+v)=cos(u)\cdot cos(v)-sin(u)\cdot sin(v)\\cos(u-v)=cos(u)\cdot cos(v)+sin(u)\cdot sin(v)\\sin(2u)=2\cdot sin(u)\cdot cos(u)\\cos(2u)=cos^2(u)-sin^2(u)\\sin^2(u)+cos^2(u)=1$

I do know the last one but I thought I'd have it here to have the whole bulk for search engines etc. in case of a good answer on the thread.

Thanks a lot!

Re: Please provide proof for the main trig functions

Those aren't simple proofs...

here you have a link of a math forum from my country with those proofs (it's in spanish :S, but you can understand if you only read the operations)

Basically, getting 1st and the 3rd and using parity and imparity of cos and sin you can demostrate the others

Funciones trigonométricas de la suma y diferencia de ángulos - Foro fmat.cl

I hope it can help you

Re: Please provide proof for the main trig functions

Quote:

Originally Posted by

**Paze** Hi MHF. If anyone has time, I would love for concrete proof for the following trig functions:

$\displaystyle sin(u+v)=sin(u)\cdot cos(v)+cos(u)\cdot sin(v)\\sin(u-v)=sin(u)\cdot cos(v)-cos(u)\cdot sin(v)\\cos(u+v)=cos(u)\cdot cos(v)-sin(u)\cdot sin(v)\\cos(u-v)=cos(u)\cdot cos(v)+sin(u)\cdot sin(v)\\sin(2u)=2\cdot sin(u)\cdot cos(u)\\cos(2u)=cos^2(u)-sin^2(u)\\sin^2(u)+cos^2(u)=1$

I do know the last one but I thought I'd have it here to have the whole bulk for search engines etc. in case of a good answer on the thread.

Thanks a lot!

2. Sin, Cos and Tan of Sum and Difference of Two Angles

As for the double angle formulas, it's simply a case of letting $\displaystyle \displaystyle \begin{align*} \beta = \alpha \end{align*}$ and simplifying.

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Re: Please provide proof for the main trig functions

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Re: Please provide proof for the main trig functions

Re: Please provide proof for the main trig functions

How you prove those things depends upon how you **define** sine and cosine.

The simplest proofs I know start from the definitions

"y= cos(x) is the function satisfying the differential equation y''= -y with initial conditions y(0)= 1, y'(0)= 0"

"y= sin(x) is the function satisfying the differential equation y''= -y with initial conditions y(0)= 0, y'(0)= 1"

It is easy to prove that sin(x) and cos(x) are independent functions so that any solution to y''= y' can be written as a linear combination of them. In fact, the solution to y''= -y, with initial conditions y(0)= A, y'(0)= B, is y(x)= Acos(x)+ Bsin(x).

If we define y(x)= (sin(x))', then y'= (sin(x))''= - sin(x) so that y''= (- sin(x))'= -y while y(0)= 1 and y'(0)= 0 so that (sin(x))'= cos(x). If we define y(x)= (cos(x))', then y'= (cos(x))''= -cos(x) so that y''= (- cos(x))'= -y while y(0)= 0 and y'(0)= -1 so that (cos(x))'= -sin(x).

If we define y(x)= sin^2(x)+ cos^2(x), we have y'= 2sin(x)cos(x)- 2cos(x)sin(x)= 0 for all x. That is, y(x) is a constant and y(0)= (sin(0))^2+ (cos(0))^2= 0^2+ 1^2= 1 so sin^2(x)+ cos^2(x)= 1 for all x.

Note that y(x)= cos(x+a) satisfies the differential equation y''= -y with initial conditions y(0)= cos(a), y'(0)= -sin(a) so that y(x)= cos(x+ a)= cos(a)cos(x)- sin(a)sin(x). Setting x= b, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

Similarly, y(x)= sin(x+ a) satisfies the differential equation y''= -y with initial conditions y(0)= sin(a), y'(0)= cos(a) so that y(x)= sin(x+a)= sin(a)cos(x)+ cos(a)sin(x). Setting x= b, cos(a+ b)= sin(a)cos(b)+ cos(a)sin(b).

You can also define $\displaystyle sin(x)= \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$, $\displaystyle cos(x)= \sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$, prove from that that sin(x) and cos(x) satisfy y''= -y with the above initial conditions and then use this proof.