sin(2x) = 4·sin(2y), show that 5·tan(x-y) = 3·tan(x+y)

We will show that both equations are equivalent to the same equation.

                                  sin(2x) = 4·sin(2y)
                          2·sin(x)·cos(x) = 4·[2sin(y)·cos(y)]
                          2·sin(x)·cos(x) = 8·sin(y)·cos(y)
                            sin(x)·cos(x) = 4·sin(y)·cos(y)
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Now we will work with the second equation:

                               5·tan(x-y) = 3·tan(x+y)

                                   = 
 
                                    = 

                       5·sin(x-y)cos(x+y) = 3·sin(x+y)cos(x-y)

5·[sin(x)cos(y)-cos(x)sin(y)][cos(x)cos(y)-sin(x)sin(y)] =
                  3·[sin(x)cos(y)+cos(x)sin(y)][cos(x)cos(y)+sin(x)sin(y)]

Since this is very bulky, I will use abbreviations:

capital S = sin(x), capital C = cos(x), small s = sin(y), small c = cos(y)


                          5[Sc-Cs][Cc-Ss] = 3[Sc+Cs][Cc+Ss]
                   5[SCc²-S²sc-C²sc+CSs²] = 3[SCc²+S²sc+C²sc+CSs²]
                   5[SCc²-(S²+C²)sc+CSs²] = 3[SCc²+(S²+C²)sc+CSs²]
                   5[SCc²+CSs²-(S²+C²)sc] = 3[SCc²+CSs²+(S²+C²)sc]
                   5[(c²+s²)SC-(S²+C²)sc] = 3[(c²+s²)SC+(S²+C²)sc]                     
                           5[(1)SC-(1)sc] = 3[(1)SC+(1)sc]
                                 5[SC-sc] = 3[SC+sc]
                                 5SC-5sc  = 3SC+3sc
                                      2SC = 8sc
                                       SC = 4sc
                            sin(x)·cos(x) = 4·sin(y)·cos(y)

So the equations are equivalent.

Edwin