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ajprincess

  • 2 years ago

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

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

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

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

  4. ajprincess
    • 2 years ago
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  5. maheshmeghwal9
    • 2 years ago
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    Nice & thanx @ajprincess :)

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

  7. alexwee123
    • 2 years ago
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    nice :)

  8. maheshmeghwal9
    • 2 years ago
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    very nice:)

  9. ajprincess
    • 2 years ago
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    Thanxxx a lot for the comments.:)

  10. experimentX
    • 2 years ago
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    seems like you used in few ,,,

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

  12. ajprincess
    • 2 years ago
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    Thanx a lot @lgbasallote

  13. WONDEMU
    • 2 years ago
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    how should I memorize these?

  14. Loujoelou
    • 2 years ago
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    I liked it :) Good Job making it :)

  15. ajprincess
    • 2 years ago
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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.

  16. apoorvk
    • 2 years ago
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    Simple, clear, lucid, and great! :P

  17. ajprincess
    • 2 years ago
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    Thanx a lot @apoorvk

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

  19. ajprincess
    • 2 years ago
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    Thanx a lot @maheshmeghwal9

  20. maheshmeghwal9
    • 2 years ago
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    :)

  21. mathslover
    • 2 years ago
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    good tutorial must appreciate ...

  22. ajprincess
    • 2 years ago
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    Thanx a lot @mathslover.

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

  24. ajprincess
    • 2 years ago
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    :)

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

  26. ajprincess
    • 2 years ago
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    thanx a lot @calcmathlete

  27. phani09
    • 2 years ago
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    that was a good revision.....

  28. ajprincess
    • 2 years ago
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    Thanx a lot @phani09

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

  30. eliassaab
    • 2 years ago
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    http://www.sosmath.com/trig/Trig5/trig5/trig5.html

  31. ajprincess
    • 2 years ago
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    ya sure. @waterineyes.

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

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

  34. Calcmathlete
    • 2 years ago
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    @waterineyes is correct. Those two identities are incorrect. \[cos(A + B) = \cos A\cos B - \sin A\sin B\]\[cos(A - B) = \cos A\cos B + \sin A\sin B\]

  35. waterineyes
    • 2 years ago
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    See, the difference between \(sin(A \pm B) \quad And \quad cos(A \pm B)\): \[\Large \color{green}{\sin(A \pm B) = sinA.cosB \pm cosA.sinB}\] \[\Large \color{green}{\cos(A \pm B) = cosA.cosB \mp cosA.cosB}\]

  36. waterineyes
    • 2 years ago
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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??

  37. ajprincess
    • 2 years ago
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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.

  38. waterineyes
    • 2 years ago
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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 ..

  39. ajprincess
    • 2 years ago
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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.

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