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Question Number 15555 by Tinkutara last updated on 11/Jun/17

Prove that  ((cos 8x − cos 7x)/(1 + 2 cos 5x)) = cos 3x − cos 2x

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{cos}\:\mathrm{8}{x}\:−\:\mathrm{cos}\:\mathrm{7}{x}}{\mathrm{1}\:+\:\mathrm{2}\:\mathrm{cos}\:\mathrm{5}{x}}\:=\:\mathrm{cos}\:\mathrm{3}{x}\:−\:\mathrm{cos}\:\mathrm{2}{x} \\ $$

Answered by ajfour last updated on 11/Jun/17

considering  f(x)=(1+2cos 5x)(cos 3x−cos 2x)    =(cos 3x−cos 2x)+2cos 5xcos 3x                           −2cos 5xcos 2x    =(cos 3x−cos 2x)+(cos 8x+cos 2x)                −(cos 7x+cos 3x)        =cos 8x−cos 7x  therefore    cos 3x−cos 2x=((cos 8x−cos 7x)/(1+2cos 5x)) .

$${considering} \\ $$$${f}\left({x}\right)=\left(\mathrm{1}+\mathrm{2cos}\:\mathrm{5}{x}\right)\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$\:\:=\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}\right)+\mathrm{2cos}\:\mathrm{5}{x}\mathrm{cos}\:\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2cos}\:\mathrm{5}{x}\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\:\:=\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}\right)+\left(\mathrm{cos}\:\mathrm{8}{x}+\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{cos}\:\mathrm{7}{x}+\mathrm{cos}\:\mathrm{3}{x}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{cos}\:\mathrm{8}{x}−\mathrm{cos}\:\mathrm{7}{x} \\ $$$${therefore}\: \\ $$$$\:\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{cos}\:\mathrm{8}{x}−\mathrm{cos}\:\mathrm{7}{x}}{\mathrm{1}+\mathrm{2cos}\:\mathrm{5}{x}}\:. \\ $$

Commented by Tinkutara last updated on 13/Jun/17

Thanks Sir but I needed to prove LHS  = RHS. Can it be done?

$$\mathrm{Thanks}\:\mathrm{Sir}\:\mathrm{but}\:\mathrm{I}\:\mathrm{needed}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{LHS} \\ $$$$=\:\mathrm{RHS}.\:\mathrm{Can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{done}? \\ $$

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