I am a little confused at to how to answer the following questions ( the ones I can't do are marked with a *).

Starting from the identity sin(A + B) = sinAcosB + cosAsinB, use the basic properties of the trig. functions (such as sin(-A) = -sinA) to prove the following:

i) sin(A - B) = sinAcosB - cosAsinB
ii)* cos(A + B) = cosAcosB - sinAsinB
iii) tan(A + B) = (tanA + tanB)/(1 - tanAtanB)
iv) sinC + sinD = 2sin0.5(C + D)cos0.5(C + D)
v)* tan^-1 a + tan^-1 b = tan^-1 ((a + b)/(1 - ab))

i) I can do this part like so,
Since sin(A + B)=sinAcosB + cosAsinB,
sub -B for +B, then we get the result we require if we remember that cos(-B)=cosB and sin(-B)=-sinB.

ii)* the question for this part says that recall that cos A = sin (p/2 - A). But I am not sure how to continue to prove this part.

iii) This part I could do. I used that we already proved sin(A + B) and cos(A + B), to say,
tan(A + B) = sin(A+B)/cos(A+B) = sinAcosB + cosAsinB / cosAcosB - sinAsinB, then divide through by cosAcosB to get the result you need.

iv) This was also ok, I remembered how to prove the factor formula.

v)* Umm. Not sure here. Is it the inverse of tan or reciprocal of tan? Can someone help out with this part? It says you can use part (iii) to help with this part.

Thanks in advance for your help.

iii) Find tan(arctan a +arctan b) and the answer pops out.
(P.S. If anyone has answered this question before me, my apologies- but I don't see any answer yet)

How can you use a double angle formula to show that cot(p/12) satisfies the equation c² - 2×3^1/2 c - 1 = 0? Then deduce cot(p/3) = 2 + 3^1/2.

My thought was to say, cot(p/12)=cot(p/6-p/12)=1-tan(p/6)tan(p/12)/tan(p/6)tan(p/12).

You have

tan(2A)=2tan(A)/(1 - tan² (A))

So

tan(p/6) = 2tan(p/12)/(1 - tan² (p/12))

Therefore

cot(p/6) = (1 - tan² (p/12))/[2tan(p/12)] = 3^1/2

So

1 - tan² (p/12) = 2×3^1/2 tan(p/12).

Then just rearrange and divide by tan² (p/12) and you're sorted.
Regards,
Olof

Thanks Olof. I am sorry for my late acknowledgement.