How do you prove that \(\displaystyle{8}{\cos{{\left({x}\right)}}}{\cos{{\left({2}{x}\right)}}}{\cos{{\left({3}{x}\right)}}}-{1}={\frac{{{\sin{{\left({7}{x}\right)}}}}}{{{\sin{{\left({x}\right)}}}}}}\) ?

Marshall Wolf

Marshall Wolf

Answered question

2022-04-10

How do you prove that 8cos(x)cos(2x)cos(3x)1=sin(7x)sin(x) ?

Answer & Explanation

WigwrannyErarmbmk

WigwrannyErarmbmk

Beginner2022-04-11Added 13 answers

We have
8sinxcos(x)cos(2x)cos(3x)=4sin(2x)cos(2x)cos(3x)
=2sin(4x)cos(3x)
Moreover
sinx+sin(7x)=2sin(x+7x2)cos(7xx2)=
and the result follows easily.

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