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Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
 one year ago
 one year ago
Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
 one year ago
 one year ago

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ajprincessBest ResponseYou've already chosen the best response.28
dw:1339728896748:dw aopposite badjacent chypotenuse. \[\sin \theta=a/c\] \[\cos \theta=b/c\] \[\tan \theta=a/b\] \[\csc \theta=1/\sin \theta=1/a/c=c/a\] \[\sec \theta=1/\cos \theta=1/b/c=c/b\] \[\cot \theta=1/\tan \theta=1/a/b=b/a\] \[\sin ^{2}\theta+ \cos ^{2}\theta=1\] \[\tan ^{2}\theta+1=\sec ^{2}\theta\] \[1+\cot^{2}\theta=\csc^{2}\theta\] \[tan\theta*\cot\theta=1\] \[sin(A+B)=sinAcosB+sinBcosA\] \[cos(A+B)=cosAcosB+sinAsinB\] \[tan(A+B)=(tanA+tanB)/(1tanA*tanB)\] \[sin(AB)=sinAcosBsinBcosA\] \[cos(AB)=cosAcosBsinAsinB\] \[tan(AB)=(tanAtanB)/(1+tanA*tanB)\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
\[sin(A+B)+sin(AB)=2sinAcosB\] \[cos(A+B)+cos(AB)=2cosAcosB\] \[sin(A+B)sin(AB)=2cosAsinB\] \[cos(AB)cos(A+B)=2sinAsinB\] \[sin2\theta=2sin\theta\cos\theta\] \[cos2\theta=cos^2\thetasin^2\theta\] \[cos2\theta=12sin^2\theta\] \[cos2\theta=2cos^2\theta1\] \[tan2\theta=2tan\theta/(1tan^2\theta)\] \[sin3\theta=3sin\theta4sin^3\theta\] \[cos3\theta=4cos^3\theta3cos\theta\] \[sin(\theta)=sin\theta\] \[cos(\theta)=cos\theta\] \[tan(theta)=tan\theta\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
\[sin(90\theta)=cos\theta\] \[cos(90\theta)=sin\theta\] \[tan(90\theta)=cot\theta\] \[sin(180\theta)=sin\theta\] \[cos(180\theta)=cos\theta\] \[tan(180\theta)=tan\theta\] \[sin(180+\theta)=sin\theta\] \[cos(180+\theta)=cos\theta\] \[tan180+\theta)=tan\theta\] \[sin(90+\theta)=cos\theta\] \[cos(90+\theta)=sin\theta\] \[tan(90+\theta)=cot\theta\]
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
Nice & thanx @ajprincess :)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
U r welcome nd thanxx. I thank all who gav me medals for givng them.:)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanxxx a lot for the comments.:)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
seems like you used in few ,,,
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
this is a true tutorial indeed \[\Huge \color{maroon}{\mathtt{\text{<tips hat>}}}\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @lgbasallote
 one year ago

WONDEMUBest ResponseYou've already chosen the best response.0
how should I memorize these?
 one year ago

LoujoelouBest ResponseYou've already chosen the best response.0
I liked it :) Good Job making it :)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @Loujoelou and as for question @wondemu memorise them group by group. For example (1)sin(A+B)=sinAcosB+sinBcosA cos(A+B)=cosAcosB+sinAsinB tan(A+B)=(tanA+tanB)/(1−tanA∗tanB) (2)sin(A−B)=sinAcosB−sinBcosA cos(A−B)=cosAcosB−sinAsinB tan(A−B)=(tanA−tanB)/(1+tanA∗tanB) You'll find them useful when u hav finishd memorising them.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
Simple, clear, lucid, and great! :P
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @apoorvk
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
Wow reaching the same target as lgba {30 medals} But hope u would get more than that ^_^
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @maheshmeghwal9
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
good tutorial must appreciate ...
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @mathslover.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
no thanks @ajprincess actually many of students present here needed this
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
Good job :) I actually didn't know the negative and triple identities.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
thanx a lot @calcmathlete
 one year ago

phani09Best ResponseYou've already chosen the best response.0
that was a good revision.....
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
Thanx a lot @phani09
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
@ajprincess , I want to tell you something if you will not feel bad.. May I tell??
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
ya sure. @waterineyes.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
Can you tell me the formulas for \(cos(A + B) \quad and \quad cos(AB)\) ??
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
@ajprincess, do you think that you have given the right identities for \(cos(A + B) \quad and \quad cos(A  B)\) ??
 one year ago

CalcmathleteBest ResponseYou've already chosen the best response.0
@waterineyes is correct. Those two identities are incorrect. \[cos(A + B) = \cos A\cos B  \sin A\sin B\]\[cos(A  B) = \cos A\cos B + \sin A\sin B\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
See, the difference between \(sin(A \pm B) \quad And \quad cos(A \pm B)\): \[\Large \color{green}{\sin(A \pm B) = sinA.cosB \pm cosA.sinB}\] \[\Large \color{green}{\cos(A \pm B) = cosA.cosB \mp cosA.cosB}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
@ajprincess , don't misunderstand me and don't think I always look for mistakes but everywhere you used this identity wrong.. So, I think your mind says it is correct but sorry to say it is incorrect... In sin(A + B) there is respective + in the middle, In sin(A  B) there is respective  in the middle.. But for cos(A + B), there is  in the middle, for cos(A  B), there is + in the middle... Getting it??
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
oh no it's k. Thanx for telling t. i copied it from a note of mine. i must hav written it wrong when I wrote. thanxx a lot for mentioning it. I welcome this sorts of comments. @waterineyes.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.0
26 people gave you medals and I think you deserve more than that.. But how will you let it know to all the 26 people that you have written slight wrong there so that they can know the real formula... Think about it.. @ajprincess ..
 one year ago

ajprincessBest ResponseYou've already chosen the best response.28
it's my mistake nd I am greatly sorry for t. Let me post t again in a new post so that they can get . I am nt sure of any other way other than this.
 one year ago
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