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# Verify identities watch

1. I am lost on this. Can someone help:

1-2cos(squared)x = sinx-cosx
_______________ ________

1-2cosxsinx sinx-cosx

you either have to make them equal to eachother or one equal to the other. that line is the divide line. and I couldnt find a better way to type squared but theres no parentheses there, just for clarification.

HELP!!!! Thanks

Josh
2. and the first bottm equation is: 1-2cosxsinx
the second bottom is: sinx-cosx

hope this helps
3. No my nightmares have returned! Trig Identities...noooooooooooooooooo oooooooooo
4. I'm sorry, but what do you want to prove? I can't understand your formatting.
5. (Original post by theone)
I'm sorry, but what do you want to prove? I can't understand your formatting.
(1-2cos²x)/(1-2 cosx sinx) == (sinx - cosx)/(sinx - cosx) [===1 ????? ]
6. (Original post by elpaw)
(1-2cos²x)/(1-2 cosx sinx) == (sinx - cosx)/(sinx - cosx) [===1 ?????]

But this isn't true.
7. (Original post by theone)
But this isn't true.
which is why i have put all them question marks
8. (Original post by elpaw)
which is why i have put all them question marks
Perhaps he means (1-2cos²x)/(1-2 cosx sinx) == (sinx + cosx)/(sinx - cosx), which I think could be true.

(1-2cos²x)/(1-2 cosx sinx) == (sinx + cosx)/(sinx - cosx)

thats it
10. I think its a sum and difference identity and the formula is either:

sin(A+B) = sinA cosB + cosA sinB

or

sin(A-B) = sinA cosB - cosA sinB

I am completely lost though and theres another one if your up for it, but that one will end in a fraction.....GULP....

josh
11. (Original post by Unregistered)

I am completely lost though and theres another one if your up for it

josh
Sure.
12. And your previous one the easiest way to go seems to be to conside the identities for Sin A - / + Cos B
13. 1 = sin²x + cos²x

=> 1 - 2 cos²x == sin²x - cos²x == (sinx + cos x)(sin x - cos x)
----------------------------------------------------------------------
1 - 2 cosx sinx == sin²x + cos²x - 2 sinx cosx == (sinx - cosx)²

(sinx - cosx) cancells from top and bottom

=> == (sinx + cosx)/(sinx - cosx)
14. Or that which is a far better method.
15. heres the other problem:

tan(x-y) -tan(y-x) == 2(tanx-tany)/1+tanx tany

I know this ends in a fraction, and it doesnt seem as hard as the other problem. I know the formula is either:

tan(A+B) == tanA+tanB/1-tanA tanB

or

tan(A-B) == tanA-tanB/1+tanA tanB

thanks guys

josh
16. it looks like you may have to work out both sides of equation to make them equal eachother. My brain hurts just thinking about it....
17. (Original post by Unregistered)
heres the other problem:

tan(x-y) -tan(y-x) == 2(tanx-tany)/1+tanx tany

I know this ends in a fraction, and it doesnt seem as hard as the other problem. I know the formula is either:

tan(A+B) == tanA+tanB/1-tanA tanB

or

tan(A-B) == tanA-tanB/1+tanA tanB

thanks guys

josh
tan(x-y) - tan(y-x) == ((tanx - tany) - (tany - tanx))/(1+ tanx tany)

== 2(tanx - tany)/(1+tanx tany)

Simple... just use the identities, like you said....
18. oh duh....I knew it was easy.

But thanks again....I bow down to almighty identity god

josh

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