1. Oct 28, 2004

dabouncerx24

Can somebody please explain to me why

sin(x+h) = sinxcosh + cosxsinh

in detail.

2. Oct 29, 2004

vsage

3. Oct 29, 2004

HallsofIvy

Staff Emeritus
That's one way.

It can also be done this way (but this is much more "mathematically sophisticated")

sin(x) and cos(x) are solutions to the differential equation y"= - y with the properties that cos(0)= 1, cos'(0)= 0, sin(0)= 1, sin'(0)= 1 so it can be shown that any solution to y"= -y with y(0)= A, y'(0)= B must be of the form y= Acos(x)+ Bsin(x).

Let y(x)= sin(x+ h) where h is a constant. Then y'(x)= cos(x+ h) and y"= -sin(x+h)= -y. This function y satisfies the differential equation. Also y(0)= sin(0+h)= sin(h) and y'(0)= cos(x+ h)= cos(x). Therefore, y(x)= sin(x+h)= sin(h)cos(x)+ cos(h)sin(x).

You can also let y(x)= cos(x+h). Then y'(x)= -sin(x+h) and y"(x)= -cos(x+h)= -y.
y(0)= cos(h) and y'(0)= -sin(h). Therefore y(x)= cos(x+h)= cos(h)cos(x)- sin(h)sin(x).

4. Oct 29, 2004

matt grime

Or use the identities

sinh(y) = {e^y - e^{-y}}/2

sin(y) = {e^{iy}-e^{-iy}}/2

etc,

if you know about complex numbers