arindameducationusc
  • arindameducationusc
New Trigonometry tutorial 1.2
Trigonometry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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arindameducationusc
  • arindameducationusc
\(\Huge\color{red}{Trigonometry} \) \(\Large\color{blue}{Trigonometry~ratios~of~the~sum~and~difference} \) \(\Large\color{blue}{of~two~angles} \) \(\Large\color{blue}{1)} \) \[\sin (A+B)=sinA cosB+cosAsinB\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{2)} \) \[\cos (A+B)=cosAcosB-sinAsinB\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{3)} \) \[\sin(A-B)=sinAcosB-cosAsinB\]

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arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{4)} \) \[\cos(A-B)=cosAcosB+sinAsinB\]
arindameducationusc
  • arindameducationusc
ya, please see the previous tutorials by me and @rvc before seeing this one..
zzr0ck3r
  • zzr0ck3r
I think we should have a tutorial section.
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{5)} \) \[\sin C+sinD=2\sin \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{6)} \) \[sinC-sinD=2\cos \frac{ C+D }{ 2 }\sin \frac{ C-D }{ 2 }\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{7)} \) \[cosC+cosD=2\cos \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]
arindameducationusc
  • arindameducationusc
\(\Large\color{blue}{7)} \) \[cosC-cosD=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{ 2 }\]
arindameducationusc
  • arindameducationusc
sorry previous was \(\Large\color{blue}{8} \)
Sepeario
  • Sepeario
very extensive and useful tutorial, ty!
arindameducationusc
  • arindameducationusc
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
Sepeario
  • Sepeario
haha nice!
rvc
  • rvc
good job !
arindameducationusc
  • arindameducationusc
\(\Large\color{red}{Proof~of=>} \) sin(A+B)=sinAcosB+cosAsinB
arindameducationusc
  • arindameducationusc
|dw:1442124539656:dw|
arindameducationusc
  • arindameducationusc
A didn't come in the figure... its right of Q |dw:1442125043892:dw|
arindameducationusc
  • arindameducationusc
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)
arindameducationusc
  • arindameducationusc
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.
arindameducationusc
  • arindameducationusc
\[=sinAcosB+cosRPNsinB\] and cosRPN=cosA so, sin(A+B)=sinAcosB+cosAsinB Hence proved!
arindameducationusc
  • arindameducationusc
More proofs for the above formulas will be in Tutorial 1.3 so, stay tuned.. Happy learning.... :)
arindameducationusc
  • arindameducationusc
@Astrophysics @Irishboy123 @Photon336 @Robert136 just check if I made any mistakes. Thank you....

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