In my text book it is directly stated that the general solution of the equation $ \sin (A) = \sin (B)$ , is
$ A = 2n\pi + B$ and $A = (2n +1)\pi - B$
but where is the second solution coming from? Since period of $\sin$ function is $2\pi$, shouldn't the solutions be, $ A = 2n\pi + B$ and $ A = 2n\pi - B$ ?
$\endgroup$4 Answers
$\begingroup$Hint :
$\sin (180-\theta)=\sin(??)$
$\endgroup$ 7 $\begingroup$say, $ \sin (A) = \sin (B) = s$ , from this we can approach the solution by finding the solution set of $ \sin (B) = s$ , since $ \sin (A) = s$ , we can say that
$ A = $ solution set of $B$. Lets solve, $ \sin (B) = s$
We will use Principal Values, Secondary Values and quadrant diagrams to approach the solution and convey clarity.
there are two possible cases, one where $s\ge 0 $ and the other where $ s \le 0$.
For the case where $s\ge 0 $, the required solution will lie in the 1st or the 2nd quadrant as shown below:
and the case where $ s \le 0$, the solution will lie in the 3rd or the 4th quadrant, as shown below:
so the general solution should give formula that will help generate the infinite set of angles. The formula(s) in question should address all the four lines, namely, PV+ , SV+, PV-, SV- and in order to generate the complete solution set for the complete range $ -1\le s \le1$
The first solution $ A = 2n\pi + B$ , is for the solutions of PV+ or PV- depending on the value of $s$. Any rotation of angles of $2n\pi$ will result in the position shown below:
Lets say the PV+ is $\theta$ and $\theta > 0$ and acute , when $ \sin (B) = s$ and $ s \ge 0$. Also lets say the PV- is $\beta$ and $\beta < 0$ and acute , when $ \sin (B) = s$ and $ s \le 0$. (Note: plugging $\beta$ into $A = 2n\pi + B$ is a subtraction as $\beta < 0$)
The diagrams below should clarify how $ A = 2n\pi + B$ gives solutions for PV+ and PV-:
(Note: plugging $\beta$ into $A = 2n\pi + B$ is a subtraction as $\beta < 0$)
Recall:
Lets say the PV+ is $\theta$ and $\theta > 0$ and acute , when $ \sin (B) = s$ and $ s \ge 0$.
Also lets say the PV- is $\beta$ and $\beta < 0$ and acute , when $ \sin (B) = s$ and $ s \le 0$.
PV or principal value is also the associated acute angle.
The second solution $A = (2n+1)\pi - B$ gives the solutions for SV+ and SV-.
any rotation of angles of $(2n+1)\pi$ will result in the position shown below:
So $ (2n+1)\pi - \theta$ gives all solution for the SV+ location, while $ (2n+1)\pi - \beta$ gives us all the solution for the SV- location.
(Note: Since $\beta<0$, $ (2n+1)\pi - \beta$ is essentially an addition)
The diagrams below should clarify how the general formula leads to SV+ and SV-:(Note: Since $\beta<0$, $ (2n+1)\pi - \beta$ is essentially an addition)
Look at the unit circle:
The y-coordinate is equal for both positions $1$ and $2$.
It remains to realise that:
Position 1) is $2n\pi +x$
Position 2) is $(2n+1)\pi-x$
EDIT:
I'm not really sure if this is what you meant. It's just an example though.
$\endgroup$ 4 $\begingroup$Equation
The general solution of the equation sin(A) = sin(B) is given by:
$A = B + (-1)^n nπ$
EDIT: Proof
Here is the proof for the above equation. we are given:
$sin(A) = sin(B)$
=> $sin(A)-sin(B)=0$
=> $2cos(\frac{A+B}{2})sin(\frac{A-B}{2}) = 0$
=> $cos(\frac{A+B}{2})=0$ or $sin(\frac{A-B}{2})=0$
=> $\frac{A+B}{2}=(2n+1)π/2$ or $\frac{A-B}{2}=nπ$
=> $A=(2n+1)π–B$ or $A=2nπ+B$
=> $A = (2n+1)π+(–1)^{2n + 1} B$ or $x = 2nπ +(–1)^{2n}B$
Combining both the equations we get,
$A = nπ + (–1)^n B$
