Always always always.wow much thanks!!
But question about second ones.
I solved only for tanx which caused me to go wrong. In which cases does both trig identifies have to be solved?
Very true. By dividing by cosx, you're assuming it can't be zero. If you choose to use this method, always check at the end whether angles like pi/2, 3pi/2 were solutions.Always always always.
The basic rule is NEVER eliminate one of the functions, because this could eliminate solutions.
e.g sinxcosx = cosx. You dont go sinx = 1 by dividing by cos x.
You would go sinxcosx-cosx = 0 and go from there.
I'm really surprised. Proving trigonometric identities at MX1 level is relatively trivial, I'd have thought.Trigonometric proofs are the absolute pits. Worse that parabola proofs, which are shocking enough.
it may be trivial but i think its extremely tedious and annoying.I'm really surprised. Proving trigonometric identities at MX1 level is relatively trivial, I'd have thought.