Question

New Trigonometry tutorial 1.2

Answer 1

\(\Huge\color{red}{Trigonometry} \)

\(\Large\color{blue}{Trigonometry~ratios~of~the~sum~and~difference} \)
\(\Large\color{blue}{of~two~angles} \)

\(\Large\color{blue}{1)} \)
\[\sin (A+B)=sinA cosB+cosAsinB\]

Answer 2

\(\Large\color{blue}{2)} \)
\[\cos (A+B)=cosAcosB-sinAsinB\]

Answer 3

\(\Large\color{blue}{3)} \)
\[\sin(A-B)=sinAcosB-cosAsinB\]

Answer 4

\(\Large\color{blue}{4)} \)
\[\cos(A-B)=cosAcosB+sinAsinB\]

Answer 5

ya, please see the previous tutorials by me and @rvc
before seeing this one..

Answer 6

http://openstudy.com/study#/updates/553f1223e4b061e2642b5eb6

and

http://openstudy.com/updates/55d8393de4b0a0d512702d8a

Answer 7

I think we should have a tutorial section.

Answer 8

\(\Large\color{blue}{5)} \)
\[\sin C+sinD=2\sin \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]

Answer 9

\(\Large\color{blue}{6)} \)
\[sinC-sinD=2\cos \frac{ C+D }{ 2 }\sin \frac{ C-D }{ 2 }\]

Answer 10

\(\Large\color{blue}{7)} \)
\[cosC+cosD=2\cos \frac{ C+D }{ 2}\cos \frac{ C-D }{ 2 }\]

Answer 11

\(\Large\color{blue}{7)} \)
\[cosC-cosD=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{ 2 }\]

Answer 12

sorry previous was \(\Large\color{blue}{8} \)

Answer 13

very extensive and useful tutorial, ty!

Answer 14

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

Answer 15

haha nice!

Answer 16

good job !

Answer 17

\(\Large\color{red}{Proof~of=>} \)
sin(A+B)=sinAcosB+cosAsinB

Answer 18

A didn't come in the figure...
its right of Q