sin(2a)=2sin(a)cos(a)
cos(2a)= cos^2(a)-sin^2(a)
cos^2(a/2)=1/2 + 1/2cos(a)
sin^2(a/2)=1/2 - 1/2cos(a)
Also, you need to know cos(a) and sin(b) (using a for alpha and b for beta)
So start there: draw a right triangle and mark one angle to be alpha. You know that the opposite side is 4 and the hypotenuse is 5, since sin(a) = o/h = 4/5 (really they could be any multiple of 4 and 5, but it won't matter in the long run.) Now, you can use the pythagorean theorem to find that the last side is 3, which means that cos(a)=3/5.
Do the same thing with b; you'll find you get a 5-12-13 triangle. So here's what we've got:
sin(a)=4/5
cos(a)=3/5
sin(b)=12/13
cos(b)=5/13
Now we can compute:
sin(2a)=2sin(a)cos(a)=2(4/5)(3/5)=24/2…
cos(2a)=cos^2(a)-sin^2(a)=9/25-16/25 = -7/25
For the 3rd, do you mean sin(a+b)? We have the sum identity
sin(a+b)=sin(a)cos(b)+sin(b)cos(a) = (4/5)(5/13) + (12/13)(3/5)
tan(2a)=sin(2a)/cos(2a), so it's 24/-7 using info from above.
cos(1/2b)= sqrt ( 1/2 + 1/2cos(b) ) = sqrt ( 1/2 + 1/2 (5/13))
tan(1/2 a) = sin(1/2 a)/cos(1/2a ), so compute cos(1/2 a) and divide by sin(1/2a) (below)
sin(1/2a) = sqrt ( 1/2 - 1/2 cos(a)) = sqrt (1/2 - 1/2 (3/5))