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C3 Trig Identity Question!

Can anyone help? Been stuck for a while...
Thanks!

Prove the identity sinx+cosx=(1-2cos^2x)/(sinx-cosx)
Original post by DonHorizon
Can anyone help? Been stuck for a while...
Thanks!

Prove the identity sinx+cosx=(1-2cos^2x)/(sinx-cosx)


Start from the RHS, try using cos2x=1sin2xcos^2x=1-sin^2x
Reply 2
Original post by NotNotBatman
Start from the RHS, try using cos2x=1sin2xcos^2x=1-sin^2x


Thanks! That's what I tried to do at first but I just can't seem to get it to work!
Original post by DonHorizon
Thanks! That's what I tried to do at first but I just can't seem to get it to work!


Could you show what you have done?
Reply 4
Original post by NotNotBatman
Could you show what you have done?


I got into a position where I had 2sin^2x-1/sinx-cosx
Reply 5
Original post by DonHorizon
I got into a position where I had 2sin^2x-1/sinx-cosx


try to get (sin^2x - cos^2x) on the top
Original post by DonHorizon
I got into a position where I had 2sin^2x-1/sinx-cosx


Okay, I've just jotted it down. Easiest way to do it is to start from the RHS and use 1=cos2x+sin2x1 = cos^2x+sin^2x , then it's straightforward from there.
Reply 7
Original post by NotNotBatman
Okay, I've just jotted it down. Easiest way to do it is to start from the RHS and use 1=cos2x+sin2x1 = cos^2x+sin^2x , then it's straightforward from there.


Ahh! I've done it now! Thanks so much!

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