Why \(\displaystyle{{\cos}^{{3}}{x}}-{2}{\cos{{\left({x}\right)}}}{{\sin}^{{2}}{\left({x}\right)}}={\frac{{{1}}}{{{4}}}}{\left({\cos{{\left({x}\right)}}}+{3}{\cos{{\left({3}{x}\right)}}}\right)}\)

sanmateo9x3

sanmateo9x3

Answered question

2022-03-12

Why
cos3x2cos(x)sin2(x)=14(cos(x)+3cos(3x))

Answer & Explanation

PahHindArrida6ee

PahHindArrida6ee

Beginner2022-03-13Added 2 answers

Here's a detailed step by step:
=14(cos(x)+3cos(3x))
=14(cos(x)+3cos(x+2x))
=14(cos(x)+3[cos(x)cos(2x)sin(x)sin(2x)])
=14(cos(x)+3[cos(x)(12sin2(x))2sin2(x)cos(x)])
=14(cos(x)+3[cos(x)4cos(x)sin2(x))])
=14(cos(x)+3cos(x)12cos(x)sin2(x))
=14(4cos(x)12cos(x)sin2(x))
=cos(x)3cos(x)sin2(x)
=cos(x)cos(x)sin2(x)2cos(x)sin2(x)
cos(x)(1sin2(x))2cos(x)sin2(x)
=cos3(x)2cos(x)sin2(x)

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