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ajprincess

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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  1. ajprincess
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    |dw:1339728896748:dw| a-opposite b-adjacent c-hypotenuse. \[\sin \theta=a/c\] \[\cos \theta=b/c\] \[\tan \theta=a/b\] \[\csc \theta=1/\sin \theta=1/a/c=c/a\] \[\sec \theta=1/\cos \theta=1/b/c=c/b\] \[\cot \theta=1/\tan \theta=1/a/b=b/a\] \[\sin ^{2}\theta+ \cos ^{2}\theta=1\] \[\tan ^{2}\theta+1=\sec ^{2}\theta\] \[1+\cot^{2}\theta=\csc^{2}\theta\] \[tan\theta*\cot\theta=1\] \[sin(A+B)=sinAcosB+sinBcosA\] \[cos(A+B)=cosAcosB+sinAsinB\] \[tan(A+B)=(tanA+tanB)/(1-tanA*tanB)\] \[sin(A-B)=sinAcosB-sinBcosA\] \[cos(A-B)=cosAcosB-sinAsinB\] \[tan(A-B)=(tanA-tanB)/(1+tanA*tanB)\]

    • one year ago
  2. ajprincess
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    \[sin(A+B)+sin(A-B)=2sinAcosB\] \[cos(A+B)+cos(A-B)=2cosAcosB\] \[sin(A+B)-sin(A-B)=2cosAsinB\] \[cos(A-B)-cos(A+B)=2sinAsinB\] \[sin2\theta=2sin\theta\cos\theta\] \[cos2\theta=cos^2\theta-sin^2\theta\] \[cos2\theta=1-2sin^2\theta\] \[cos2\theta=2cos^2\theta-1\] \[tan2\theta=2tan\theta/(1-tan^2\theta)\] \[sin3\theta=3sin\theta-4sin^3\theta\] \[cos3\theta=4cos^3\theta-3cos\theta\] \[sin(-\theta)=-sin\theta\] \[cos(-\theta)=cos\theta\] \[tan(-theta)=-tan\theta\]

    • one year ago
  3. ajprincess
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    \[sin(90-\theta)=cos\theta\] \[cos(90-\theta)=sin\theta\] \[tan(90-\theta)=cot\theta\] \[sin(180-\theta)=sin\theta\] \[cos(180-\theta)=-cos\theta\] \[tan(180-\theta)=-tan\theta\] \[sin(180+\theta)=-sin\theta\] \[cos(180+\theta)=-cos\theta\] \[tan180+\theta)=tan\theta\] \[sin(90+\theta)=cos\theta\] \[cos(90+\theta)=-sin\theta\] \[tan(90+\theta)=-cot\theta\]

    • one year ago
  4. ajprincess
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    • one year ago
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  5. maheshmeghwal9
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    Nice & thanx @ajprincess :)

    • one year ago
  6. ajprincess
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    U r welcome nd thanxx. I thank all who gav me medals for givng them.:)

    • one year ago
  7. alexwee123
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    nice :)

    • one year ago
  8. maheshmeghwal9
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    very nice:)

    • one year ago
  9. ajprincess
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    Thanxxx a lot for the comments.:)

    • one year ago
  10. experimentX
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    seems like you used in few ,,,

    • one year ago
  11. lgbasallote
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    this is a true tutorial indeed \[\Huge \color{maroon}{\mathtt{\text{<tips hat>}}}\]

    • one year ago
  12. ajprincess
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    Thanx a lot @lgbasallote

    • one year ago
  13. WONDEMU
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    how should I memorize these?

    • one year ago
  14. Loujoelou
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    I liked it :) Good Job making it :)

    • one year ago
  15. ajprincess
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    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
  16. apoorvk
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    Simple, clear, lucid, and great! :P

    • one year ago
  17. ajprincess
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    Thanx a lot @apoorvk

    • one year ago
  18. maheshmeghwal9
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    Wow reaching the same target as lgba {30 medals} But hope u would get more than that ^_^

    • one year ago
  19. ajprincess
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    Thanx a lot @maheshmeghwal9

    • one year ago
  20. maheshmeghwal9
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    :)

    • one year ago
  21. mathslover
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    good tutorial must appreciate ...

    • one year ago
  22. ajprincess
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    Thanx a lot @mathslover.

    • one year ago
  23. mathslover
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    no thanks @ajprincess actually many of students present here needed this

    • one year ago
  24. ajprincess
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    :)

    • one year ago
  25. Calcmathlete
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    Good job :) I actually didn't know the negative and triple identities.

    • one year ago
  26. ajprincess
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    thanx a lot @calcmathlete

    • one year ago
  27. phani09
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    that was a good revision.....

    • one year ago
  28. ajprincess
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    Thanx a lot @phani09

    • one year ago
  29. waterineyes
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    @ajprincess , I want to tell you something if you will not feel bad.. May I tell??

    • one year ago
  30. eliassaab
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    http://www.sosmath.com/trig/Trig5/trig5/trig5.html

    • one year ago
  31. ajprincess
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    ya sure. @waterineyes.

    • one year ago
  32. waterineyes
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    Can you tell me the formulas for \(cos(A + B) \quad and \quad cos(A-B)\) ??

    • one year ago
  33. waterineyes
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    @ajprincess, do you think that you have given the right identities for \(cos(A + B) \quad and \quad cos(A - B)\) ??

    • one year ago
  34. Calcmathlete
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    @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
  35. waterineyes
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    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
  36. waterineyes
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    @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
  37. ajprincess
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    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
  38. waterineyes
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    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
  39. ajprincess
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    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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