[Jez]

Commitment

The whole Cos, Sin and Tan of Trig is fine - especially when used for real, with triangles.

Just you wait till you (might) have to work with algebraic expressions feating Cos^2 and (SinTan)^-2

(I gave up on it and just blagged it during the exam. Got a C by the end of that year, which I was happy with (AS Level Maths (couldn't face a second year of it )

Example (from a random site, which happened to be the BBC as that was the SE I used!):

Question:-
use any standard identies to prove the following:
a) cos(A+B).cos(A-B) = cos^2A - sin^B
b) sinx / 1+cosx = tan(x/2)

Answer:-

This answer is posted on behalf of Nicholas Dean.
a) The identities
cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
will enable you to express cos(A+B)cos(A-B) in terms of cosA, cosB, sinA and sinB. Then use
cos2 + sin2 = 1 (=> cos2 = 1 - sin2 , etc.)
to obtain an expression in terms of cosA and sinB, i.e. cos2A - sin2B.

b) The identity
for t = tan(A/2): sinA = 2t/(1+t2), cosA = (1-t2)/( 1+t2)
will convert all the functions into terms of t. The simplification of the fraction which is the left-hand side of the equation is fairly straightforward; if you cannot obtain the result, write down each step (multiply top and bottom of the fraction by (1+ t2), etc.) of your calculations.


....nope, me neither....but at least I went over my 10 char limit....albeit with my initial word!