Trig Proof (1 Viewer)

[MC Squidge]

find the general solution for sin[3x]sin[x]=2cos[2x]+1

the hint was prove 1/2(cos[(M-N)x]-cos[(M+N)x]=sin[Mx]sin[Nx]

 

[abc123doremi]

LHS=
sin 3x . sin x = -0.5 [cos 4x - cos x]
= -0.5 [(2cos²2x-1) - cos x]
= -0.5 [(2{2cosx -1}² -1 - cos x]

RHS =2 (2cos²x -1) +1

equate LHS=RHS to form a quadratic in cos x


i haven't brushed on my double-angle formulae etc so i might be wrong
but you get the idea


oh and the proof they require of you is simple ...just expand using addition formulae. that's what i used in the first lines of this proof

 

[study-freak]

 

[MC Squidge]

thanks