Prove that \(\displaystyle{\frac{{{\cos{{3}}}{A}+{\cos{{3}}}{B}}}{{{2}{\cos{{\left({A}-{B}\right)}}}-{1}}}}={\left({\cos{{A}}}+{\cos{{B}}}\right)}{\cos{{\left({A}+{B}\right)}}}-{\left({\sin{{A}}}+{\sin{{B}}}\right)}{\sin{{\left({A}+{B}\right)}}}\) I used \(\displaystyle{\cos{{3}}}{A}={4}{{\cos}^{{3}}{A}}-{3}{\cos{{A}}}\), but

markush35q

markush35q

Answered question

2022-04-01

Prove that cos3A+cos3B2cos(AB)1=(cosA+cosB)cos(A+B)(sinA+sinB)sin(A+B)
I used cos3A=4cos3A3cosA, but it is getting more and more complicated.

Answer & Explanation

Nunnaxf18

Nunnaxf18

Beginner2022-04-02Added 18 answers

(cosα+cosβ)cos(α+β)(sinα+sinβ)sin(α+β)=
cosαcos(α+β)sinαsin(α+β)+cosβcos(α+β)sinβsin(α+β)=
=cos(2α+β)+cos(2β+α)=
=2cos3α+3β2cosαβ2
and
cos3α+cos3β2cos(αβ)1=2cos3α+3β2cos3α3β22cos(αβ)1
Thus, it remains to prove that
cos3α3β2=cosαβ2(2cos(αβ)1),
which is true because
cos3α3β2=4cos3αβ23cosαβ2=
=cosαβ2(2(1+cos(αβ))3)=
=cosαβ2(2cos(αβ)1)
Done

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