# Double angle proof's

1. Sep 3, 2011

### Hogart

Hello,

I'm currently trying to convert polar equations to Cartesian equations and vise-versa. I noticed an equation needed a Trig ID I am not familiar with:

sin(2theta) = 2sin(theta)*cos(theta)

cos(2theta) = cos^2(theta) - sin^2(theta)

My apologies if that looks like an eye soar; I'm new here and don't know how to use those fancy characters. Anyway, I don't recall these ID's. I might have not been paying attention in Trig.

If anyone can explain to me how they are equal I would greatly appreciate it!

2. Sep 3, 2011

### BloodyFrozen

Use the addition (or if you want subtraction) identity.

sin( a+b ) = sin(a)cos(b) + cos(a)sin(b)
cos( a+b ) = cos(a)cos(b) - sin(a)sin(b)

Setting a for theta and b also for theta. See if you can figure it out...

sin( theta+theta ) = ??
cos( theta+theta) = ??

3. Sep 4, 2011

### stallionx

If a vector is r*Cos(A) i + r*sin(A) j

and the other is :

r*Cos(B) i + r*sin(B) j

and you dot product them

and A is bigger than B by A-B difference

r**2 cos A cos B + r**2 Sin A Sin B = r**2 cos( angle between)

cos A cos B + sin A sin B = cos(A-B) if B=-B

it is easy toı show that

Cos(A+B) = cosA cos B - Sin A sin B

And where sin**2 + cos**2 = 1

you can find

Sin(A+B)=Sin A cosB + cos A sin B

Last edited: Sep 4, 2011
4. Sep 4, 2011

### stallionx

Also Cos(90 - (A+B)) = Sin(A+B)
cos((90-A ) - B)
Apply result from proof

cos (90-A) cos(B)+sin (90-A) sin(B) = Sin(A+B)

SinA cosB+cosA sin B = Sin(A+B)