Can't seem to find the markscheme (official or unofficial anywhere ;/!)
Can anyone help me on this question?
[sin2x]/[1 + cos2x] = tanx
I chose to start with the LHS and so far have got;
[2sinxcosx]/[1 + cosx]
and now not sure what to do
C3 solving problem - stuck, help :(! Watch
- Thread Starter
- 28-12-2010 18:12
- 28-12-2010 22:07
I take it that 1+cosx is a typo and you meant [2sinxcosx]/[1 + cos2x], so then you can replace cos2x with cos²x-sin²x which is a double angle formula i think, So then you have [2sinxcosx]/[1 + cos²x-sin²x] and as: cos²x=1-sin²x you can sub this in to get [2sinxcosx]/[2cos²x] then cancel out a cosx and the 2s and you got tanx.Last edited by Rjp; 28-12-2010 at 22:08.
Last edited by implus; 30-12-2010 at 13:56.
- 30-12-2010 13:44
- 30-12-2010 13:51
cos2x = (cosx)^2 - (sinx)^2
1 = (sinx)^2 + (cosx)^2
when you add them together:
(sinx)^2 + (cosx)^2 + (cosx)^2 - (sinx)^2 = 2(cosx)^2
Now we are left with:
We know sin2x = 2sinxcosx
Hence you are left with sinx/cosx