# Trig proof website

1. Sep 26, 2004

### quasar987

Does someone know of a website that has proofs for most basic trigonometry identities?

for sin(x+y), cos(x+y), sin(2x), cos(2x), tg²x + 1 = sec²x, etc

thanks!

2. Sep 27, 2004

### pig

You can use the Euler formula:
cos(x) + isin(x) = e^(ix)
So:
cos(x+y) + isin(x+y) = e^[i(x+y)]
cos(x+y) + isin(x+y) = e^(ix+iy)
cos(x+y) + isin(x+y) = e^ix * e^iy
Turn the right side into sines and cosines using the original formula and see what you get.

This way you can prove the additive formulas, and the other identities like sin(2x) or cos(x)+cos(y) or tg(x+y) can be proven using them. I don't know about the last one you mentioned since I have no idea what sec(x) is.

3. Sep 27, 2004

### Muzza

sec(x) is 1/cos(x). The last identity follows very easily from the Pythagorean identity, sin^2(x) + cos^2(x) = 1.

4. Sep 27, 2004

### HallsofIvy

Staff Emeritus
That depends strongly on how you define sine and cosine.

If you define them by the "elementary" right triangle ratios, sin2x+ cos2x= 1 follows from the Pythagorean theorem and tan2 x+ 1= sec2 x follows from that. However, such things as sin(x+y), cos(x+y) etc. may not even make sense.

If you define them by sin(x)= (eix- e-ix)/(2i) and cos(x)= (eix+eix)/2, then pig's method can be used.

If you define them by "sin(x) is the function, y, satisfying y"= -y, y(0)= 0, y'(0)= 1" and "cos(x) is the function, y, satisfying y"= -y, y(0)= 1, y'(0)= 0"
Then you can show that any function satisfying y"= -y, y(0)= a, y'(0)= b must be y= a cos(x)+ b sin(x). In particular, for example, cos(x+a) satisfies
y"= -y, y(0)= cos(a), y'(0)= -sin(a) so cos(x+a)= cos(a)cos(x)- sin(a)sin(x) and, with x= b, cos(a+b)= cos(a)cos(b)- sin(a)sin(b).