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AmTran_Bus
 4 years ago
Trig identities...harder one
AmTran_Bus
 4 years ago
Trig identities...harder one

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AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353627728354:dw

hba
 4 years ago
Best ResponseYou've already chosen the best response.1Try this. First, change all tan(x) and cot(x) into sin(x)/cos(x) and cos(x)/sin(x) respectively. Then combine the fractions using common denominator which is cos(x)sin(x). Hba.

hba
 4 years ago
Best ResponseYou've already chosen the best response.1@AmTran_Bus Try it and tell me what you get ?

AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0I kept getting it to equal sin2. Let me give it another try.

hba
 4 years ago
Best ResponseYou've already chosen the best response.1No if you try to reduce it,It will come to \[ \sin^2x \cos^2x \]

AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0Agreed. I see you now. I have it on my paper. But now you have your +2cos^2 on there. I know sin+cos=1 (as long as they are both squared)...but what about the rest?

hba
 4 years ago
Best ResponseYou've already chosen the best response.1Then put it in the actual ques. \[\sin^2x \cos^2x + 2\cos^2x = 1\]

hba
 4 years ago
Best ResponseYou've already chosen the best response.1Now you know how it goes ? @AmTran_Bus

AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0I understand that, yes. But maybe I just can't see it because I can't carry it on.

AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1353628964067:dw

AmTran_Bus
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yes. Thanks and Happy Thanksgiving.
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