Solving Sin(A)=Sin(B): Where did the 2nd Solution Go?

[C_Ovidiu]

Say we have the ecuations below a,b,c positive numbers
sin(A)=a/sqrt(a^2+b^2) 1
sin(B)=a/sqrt(a^2+c^2) 2
sin(A)cos(A)=sin(B)cos(B) 3

I realized that calculating cos(A) and cos(B) and then solving the third ecuations i get the solutions b=c and a^2=bc
But solving only sin(A)=sin(B) ( we get form 3 that A=B ) I get only b=c . I'd like to know where did the second solution go ... ??

 

[sadhu]

did anyone suggested you of latex format

 

[C_Ovidiu]

Nobody ( just) did

 

[Barking_Mad]

sin(A)\ =\ \frac{a}{\left(a^{2}+b^{2}\right)^{\frac{1}{2}}}

sin(B)\ =\ \frac{a}{\left(a^{2}+c^{2}\right)^{\frac{1}{2}}}

sin(A) \times cos(A)\ =\ sin(B) \times cos(B)

Did it for you :P

 

[C_Ovidiu]

Tks. I shall learn this latex .

 

[dodo]

It is not really true that b=c, or that A=B. They could be equal, but there are many solutions where they are different, for example a=sqrt(3), b=1, c=3, A=60 deg, B=30 deg.

If you draw a plot of the function f(x) = sin(x)cos(x), it could help you to visualize the relation between A and B (and thus between b and c).

Also, since a,b,c > 0, you can draw them as line segments, so it might help to draw a picture with triangles that represent eq. 1 and 2 (and where b is not necessarily equal to c).

 

[sadhu]

one thing is clear seeing the eq. they certainly represent the right triangle with a perpendicular dropped on hypotensue

a=length of perpendicular
b,c sides

i hope their is no specific solution for the angles

they are just complementary

don,t go by it it is just a guess , no calculations