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ajprincess
Group Title
Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
 2 years ago
 2 years ago
ajprincess Group Title
Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
 2 years ago
 2 years ago

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

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

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
\[sin(90\theta)=cos\theta\] \[cos(90\theta)=sin\theta\] \[tan(90\theta)=cot\theta\] \[sin(180\theta)=sin\theta\] \[cos(180\theta)=cos\theta\] \[tan(180\theta)=tan\theta\] \[sin(180+\theta)=sin\theta\] \[cos(180+\theta)=cos\theta\] \[tan180+\theta)=tan\theta\] \[sin(90+\theta)=cos\theta\] \[cos(90+\theta)=sin\theta\] \[tan(90+\theta)=cot\theta\]
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
Nice & thanx @ajprincess :)
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
U r welcome nd thanxx. I thank all who gav me medals for givng them.:)
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
very nice:)
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanxxx a lot for the comments.:)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
seems like you used in few ,,,
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
this is a true tutorial indeed \[\Huge \color{maroon}{\mathtt{\text{<tips hat>}}}\]
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanx a lot @lgbasallote
 2 years ago

WONDEMU Group TitleBest ResponseYou've already chosen the best response.0
how should I memorize these?
 2 years ago

Loujoelou Group TitleBest ResponseYou've already chosen the best response.0
I liked it :) Good Job making it :)
 2 years ago

ajprincess Group TitleBest 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.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Simple, clear, lucid, and great! :P
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanx a lot @apoorvk
 2 years ago

maheshmeghwal9 Group TitleBest 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 ^_^
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanx a lot @maheshmeghwal9
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
good tutorial must appreciate ...
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanx a lot @mathslover.
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
no thanks @ajprincess actually many of students present here needed this
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
Good job :) I actually didn't know the negative and triple identities.
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
thanx a lot @calcmathlete
 2 years ago

phani09 Group TitleBest ResponseYou've already chosen the best response.0
that was a good revision.....
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
Thanx a lot @phani09
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
@ajprincess , I want to tell you something if you will not feel bad.. May I tell??
 2 years ago

eliassaab Group TitleBest ResponseYou've already chosen the best response.0
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
 2 years ago

ajprincess Group TitleBest ResponseYou've already chosen the best response.28
ya sure. @waterineyes.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.0
Can you tell me the formulas for \(cos(A + B) \quad and \quad cos(AB)\) ??
 2 years ago

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

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
@waterineyes is correct. Those two identities are incorrect. \[cos(A + B) = \cos A\cos B  \sin A\sin B\]\[cos(A  B) = \cos A\cos B + \sin A\sin B\]
 2 years ago

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

waterineyes Group TitleBest 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??
 2 years ago

ajprincess Group TitleBest 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.
 2 years ago

waterineyes Group TitleBest 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 ..
 2 years ago

ajprincess Group TitleBest 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.
 2 years ago
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