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arindameducationusc
 one year ago
New Trigonometry tutorial 1.2
arindameducationusc
 one year ago
New Trigonometry tutorial 1.2

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Huge\color{red}{Trigonometry} \) \(\Large\color{blue}{Trigonometry~ratios~of~the~sum~and~difference} \) \(\Large\color{blue}{of~two~angles} \) \(\Large\color{blue}{1)} \) \[\sin (A+B)=sinA cosB+cosAsinB\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{2)} \) \[\cos (A+B)=cosAcosBsinAsinB\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{3)} \) \[\sin(AB)=sinAcosBcosAsinB\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{4)} \) \[\cos(AB)=cosAcosB+sinAsinB\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8ya, please see the previous tutorials by me and @rvc before seeing this one..

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8http://openstudy.com/study#/updates/553f1223e4b061e2642b5eb6 and http://openstudy.com/updates/55d8393de4b0a0d512702d8a

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I think we should have a tutorial section.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{5)} \) \[\sin C+sinD=2\sin \frac{ C+D }{ 2}\cos \frac{ CD }{ 2 }\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{6)} \) \[sinCsinD=2\cos \frac{ C+D }{ 2 }\sin \frac{ CD }{ 2 }\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{7)} \) \[cosC+cosD=2\cos \frac{ C+D }{ 2}\cos \frac{ CD }{ 2 }\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{blue}{7)} \) \[cosCcosD=2\sin \frac{ C+D }{ 2 }\sin \frac{ DC }{ 2 }\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8sorry previous was \(\Large\color{blue}{8} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0very extensive and useful tutorial, ty!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8Thank 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

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\(\Large\color{red}{Proof~of=>} \) sin(A+B)=sinAcosB+cosAsinB

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8dw:1442124539656:dw

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8A didn't come in the figure... its right of Q dw:1442125043892:dw

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8Let 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
 one year ago
Best ResponseYou've already chosen the best response.8Hence, \[\sin(A+B)=sinAOP=\frac{ MP }{ OP }=\frac{ MR+RP }{ OP } \] \[\frac{ QN }{ OP }+\frac{ RP }{ OP }=\frac{ QN }{ ON }\frac{ ON }{ OP }+\frac{ RP }{ NP }\frac{ NP }{OP }\] as QN=MR.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8\[=sinAcosB+cosRPNsinB\] and cosRPN=cosA so, sin(A+B)=sinAcosB+cosAsinB Hence proved!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8More proofs for the above formulas will be in Tutorial 1.3 so, stay tuned.. Happy learning.... :)

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.8@Astrophysics @Irishboy123 @Photon336 @Robert136 just check if I made any mistakes. Thank you....
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