- #1
stefanB
- 9
- 0
Hello, I have been trying to solve this type of question but the results of my efforts are frequently disappointing. Since this is considered elementary and simple I have no doubt that someone can help me with this:
Solve arcsin (3/5) + arctan (1/7).
What if it had said 641/19 instead of 3/5, and 3/711 instead of 1/7? My teachers guided me towards:
tan x = tan (arcsin 3/5 + arctan 1/7) =, if arcsin 3/5 = A and arctan 1/7 = B, =
sin(A + B)/cos(A + B) which I develop to [(sinAcosB + cosAsinB)/(cosAcosB + sinAsinB)].
I do not know what to do next, and I certainly do not see how this is elementary - or am I trying to solve it the wrong way? I know the answer but the method eludes me.
I apologize if above looks messy; some posts I have read contains very structured mathematical language and it seems to be in a different font - larger and bolder, but I don't know how to write that.
Thanks in advance.
Solve arcsin (3/5) + arctan (1/7).
What if it had said 641/19 instead of 3/5, and 3/711 instead of 1/7? My teachers guided me towards:
tan x = tan (arcsin 3/5 + arctan 1/7) =, if arcsin 3/5 = A and arctan 1/7 = B, =
sin(A + B)/cos(A + B) which I develop to [(sinAcosB + cosAsinB)/(cosAcosB + sinAsinB)].
I do not know what to do next, and I certainly do not see how this is elementary - or am I trying to solve it the wrong way? I know the answer but the method eludes me.
I apologize if above looks messy; some posts I have read contains very structured mathematical language and it seems to be in a different font - larger and bolder, but I don't know how to write that.
Thanks in advance.