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arindameducationusc

  • one year ago

New Trigonometry tutorial 1.2

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  1. arindameducationusc
    • one year ago
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    \(\Huge\color{red}{Trigonometry} \) \(\Large\color{blue}{Trigonometry~ratios~of~the~sum~and~difference} \) \(\Large\color{blue}{of~two~angles} \) \(\Large\color{blue}{1)} \) \[\sin (A+B)=sinA cosB+cosAsinB\]

  2. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{2)} \) \[\cos (A+B)=cosAcosB-sinAsinB\]

  3. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{3)} \) \[\sin(A-B)=sinAcosB-cosAsinB\]

  4. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{4)} \) \[\cos(A-B)=cosAcosB+sinAsinB\]

  5. arindameducationusc
    • one year ago
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    ya, please see the previous tutorials by me and @rvc before seeing this one..

  6. zzr0ck3r
    • one year ago
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    I think we should have a tutorial section.

  7. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{5)} \) \[\sin C+sinD=2\sin \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]

  8. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{6)} \) \[sinC-sinD=2\cos \frac{ C+D }{ 2 }\sin \frac{ C-D }{ 2 }\]

  9. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{7)} \) \[cosC+cosD=2\cos \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]

  10. arindameducationusc
    • one year ago
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    \(\Large\color{blue}{7)} \) \[cosC-cosD=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{ 2 }\]

  11. arindameducationusc
    • one year ago
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    sorry previous was \(\Large\color{blue}{8} \)

  12. sepeario
    • one year ago
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    very extensive and useful tutorial, ty!

  13. arindameducationusc
    • one year ago
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    Thank you @Sepeario I will prove some of the above tomorrow, don't worry. I will have to take rest, tomorrow I am running a marathon

  14. sepeario
    • one year ago
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    haha nice!

  15. rvc
    • one year ago
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    good job !

  16. arindameducationusc
    • one year ago
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    \(\Large\color{red}{Proof~of=>} \) sin(A+B)=sinAcosB+cosAsinB

  17. arindameducationusc
    • one year ago
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    |dw:1442124539656:dw|

  18. arindameducationusc
    • one year ago
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    A didn't come in the figure... its right of Q |dw:1442125043892:dw|

  19. arindameducationusc
    • one year ago
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    Let the revolving line start from OA and trace out the angle AOB, and then trace out further angle BOC(=B). In the final position of the revolving line take any point P, and draw PM and PN perpendicular to OA and OB respectively; through N draw NR parallel to AO to meet MP in R, and draw NQ perpendicular to OA. The angle, \[RPN=90^{o}-∠PNR=∠RNO=∠NOQ=A\] (if you can't understand how, ask me)

  20. arindameducationusc
    • one year ago
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    Hence, \[\sin(A+B)=sinAOP=\frac{ MP }{ OP }=\frac{ MR+RP }{ OP } \] \[\frac{ QN }{ OP }+\frac{ RP }{ OP }=\frac{ QN }{ ON }\frac{ ON }{ OP }+\frac{ RP }{ NP }\frac{ NP }{OP }\] as QN=MR.

  21. arindameducationusc
    • one year ago
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    \[=sinAcosB+cosRPNsinB\] and cosRPN=cosA so, sin(A+B)=sinAcosB+cosAsinB Hence proved!

  22. arindameducationusc
    • one year ago
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    More proofs for the above formulas will be in Tutorial 1.3 so, stay tuned.. Happy learning.... :)

  23. arindameducationusc
    • one year ago
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    @Astrophysics @Irishboy123 @Photon336 @Robert136 just check if I made any mistakes. Thank you....

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