omg I hate these things...
1. sin^4A + 2cos^2A - cos^4A = 1
2. cos3A = -3cosA+4cos^3A
3. cos^4A = 1/8(3 + 4cos2A + cos4A)
4. (1 - cosA)/sinA = tan(A/2)
5. sin2A = (2tanA)/1 + tan^2A
Well, these are all of them so far, because I haven't looked at them all yet. Trust me I tried to do these by myself, and succeded in some not on here, but I'm not an identities person.
If you can only post about one or two problems per posting, then these are inviting problems. More than two are causing loss of interest.
These 5 problems, you should have separated them into 5 , 4, or 3 postings.
So I try the first two here. If you can re-post the next 3 in 3 or 2 separate postings, then I'd try to solve them also.
--------------------
1) sin^4A + 2cos^2A - cos^4A = 1
LHS =
= sin^4(A) +2cos^2(A) -cos^4(A)
Rearranging,
= [sin^4(A) -cos^4(A)] +2cos^2(A)
= [(sin^2(A) +cos^2(A))*(sin^2(A) -cos^2(A))] +2cos^2(A)
= [(1)*(sin^2(A) -cos^2(A))] +2cos^2(A)
= sin^2(A) -cos^2(A) +2cos^2(A)
= sin^2(A) +cos^2(A)
= 1
= RHS
Therefore, proven.
Note: We used,
----sin^2(A) +cos^2(A) = 1
----a^2 -b^2 = (a+b)(a-b)
-----------------------------------------------
2) cos3A = -3cosA+4cos^3A
LHS =
= cos(3A)
= cos(2A +A)
= cos(2A)cosA -sin(2A)sinA
= [2cos^2(A) -1]cosA -[2sinAcosA]sinA
= [2cos^3(A) -cosA] -2sin^2(A)cosA
= 2cos^3(A) -cosA -2cosA[1 -cos^2(A)]
= 2cos^3(A) -cosA -2cosA +2cos^3(A)
= -3cosA +4cos^3(A)
= RHS
Therefore, proven.
We used:
----cos(A+B) = cosAcosB -sinAsinB
----cos(2A) = 2cos^2(A) -1
----sin(2A) = 2sinAcosA
----sin^2(A) = 1 -cos^2(A)
They're not. We have some identites but not these ones. All we have is 8 of the easiest ones. Anything multiplied or divided by two, or even associated with the angles is in the AP or Pre-calculus course. This is Canada for you...I think you made a mistake these identities must be on you exam. They appear in all textbooks I seen.