ajprincess
  • ajprincess
Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
Mathematics
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ajprincess
  • ajprincess
Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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ajprincess
  • ajprincess
|dw:1339728896748:dw| a-opposite b-adjacent c-hypotenuse. \[\sin \theta=a/c\] \[\cos \theta=b/c\] \[\tan \theta=a/b\] \[\csc \theta=1/\sin \theta=1/a/c=c/a\] \[\sec \theta=1/\cos \theta=1/b/c=c/b\] \[\cot \theta=1/\tan \theta=1/a/b=b/a\] \[\sin ^{2}\theta+ \cos ^{2}\theta=1\] \[\tan ^{2}\theta+1=\sec ^{2}\theta\] \[1+\cot^{2}\theta=\csc^{2}\theta\] \[tan\theta*\cot\theta=1\] \[sin(A+B)=sinAcosB+sinBcosA\] \[cos(A+B)=cosAcosB+sinAsinB\] \[tan(A+B)=(tanA+tanB)/(1-tanA*tanB)\] \[sin(A-B)=sinAcosB-sinBcosA\] \[cos(A-B)=cosAcosB-sinAsinB\] \[tan(A-B)=(tanA-tanB)/(1+tanA*tanB)\]
ajprincess
  • ajprincess
\[sin(A+B)+sin(A-B)=2sinAcosB\] \[cos(A+B)+cos(A-B)=2cosAcosB\] \[sin(A+B)-sin(A-B)=2cosAsinB\] \[cos(A-B)-cos(A+B)=2sinAsinB\] \[sin2\theta=2sin\theta\cos\theta\] \[cos2\theta=cos^2\theta-sin^2\theta\] \[cos2\theta=1-2sin^2\theta\] \[cos2\theta=2cos^2\theta-1\] \[tan2\theta=2tan\theta/(1-tan^2\theta)\] \[sin3\theta=3sin\theta-4sin^3\theta\] \[cos3\theta=4cos^3\theta-3cos\theta\] \[sin(-\theta)=-sin\theta\] \[cos(-\theta)=cos\theta\] \[tan(-theta)=-tan\theta\]
ajprincess
  • ajprincess
\[sin(90-\theta)=cos\theta\] \[cos(90-\theta)=sin\theta\] \[tan(90-\theta)=cot\theta\] \[sin(180-\theta)=sin\theta\] \[cos(180-\theta)=-cos\theta\] \[tan(180-\theta)=-tan\theta\] \[sin(180+\theta)=-sin\theta\] \[cos(180+\theta)=-cos\theta\] \[tan180+\theta)=tan\theta\] \[sin(90+\theta)=cos\theta\] \[cos(90+\theta)=-sin\theta\] \[tan(90+\theta)=-cot\theta\]

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ajprincess
  • ajprincess
1 Attachment
maheshmeghwal9
  • maheshmeghwal9
Nice & thanx @ajprincess :)
ajprincess
  • ajprincess
U r welcome nd thanxx. I thank all who gav me medals for givng them.:)
alexwee123
  • alexwee123
nice :)
maheshmeghwal9
  • maheshmeghwal9
very nice:)
ajprincess
  • ajprincess
Thanxxx a lot for the comments.:)
experimentX
  • experimentX
seems like you used in few ,,,
lgbasallote
  • lgbasallote
this is a true tutorial indeed \[\Huge \color{maroon}{\mathtt{\text{}}}\]
ajprincess
  • ajprincess
Thanx a lot @lgbasallote
anonymous
  • anonymous
how should I memorize these?
anonymous
  • anonymous
I liked it :) Good Job making it :)
ajprincess
  • ajprincess
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.
apoorvk
  • apoorvk
Simple, clear, lucid, and great! :P
ajprincess
  • ajprincess
Thanx a lot @apoorvk
maheshmeghwal9
  • maheshmeghwal9
Wow reaching the same target as lgba {30 medals} But hope u would get more than that ^_^
ajprincess
  • ajprincess
Thanx a lot @maheshmeghwal9
maheshmeghwal9
  • maheshmeghwal9
:)
mathslover
  • mathslover
good tutorial must appreciate ...
ajprincess
  • ajprincess
Thanx a lot @mathslover.
mathslover
  • mathslover
no thanks @ajprincess actually many of students present here needed this
ajprincess
  • ajprincess
:)
anonymous
  • anonymous
Good job :) I actually didn't know the negative and triple identities.
ajprincess
  • ajprincess
thanx a lot @calcmathlete
anonymous
  • anonymous
that was a good revision.....
ajprincess
  • ajprincess
Thanx a lot @phani09
anonymous
  • anonymous
@ajprincess , I want to tell you something if you will not feel bad.. May I tell??
anonymous
  • anonymous
http://www.sosmath.com/trig/Trig5/trig5/trig5.html
ajprincess
  • ajprincess
ya sure. @waterineyes.
anonymous
  • anonymous
Can you tell me the formulas for \(cos(A + B) \quad and \quad cos(A-B)\) ??
anonymous
  • anonymous
@ajprincess, do you think that you have given the right identities for \(cos(A + B) \quad and \quad cos(A - B)\) ??
anonymous
  • anonymous
@waterineyes is correct. Those two identities are incorrect. \[cos(A + B) = \cos A\cos B - \sin A\sin B\]\[cos(A - B) = \cos A\cos B + \sin A\sin B\]
anonymous
  • anonymous
See, the difference between \(sin(A \pm B) \quad And \quad cos(A \pm B)\): \[\Large \color{green}{\sin(A \pm B) = sinA.cosB \pm cosA.sinB}\] \[\Large \color{green}{\cos(A \pm B) = cosA.cosB \mp cosA.cosB}\]
anonymous
  • anonymous
@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??
ajprincess
  • ajprincess
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.
anonymous
  • anonymous
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 ..
ajprincess
  • ajprincess
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.

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