prove that sin 3x *sin cube x *+ cos 3x * cos cube x = cos cube 2x

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$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=\sin \left(3\mathrm{x}\right).{\sin }^3\mathrm{x}+\cos \left(3\mathrm{x}\right).{\cos }^3\mathrm{x} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \sin \left(3\mathrm{x}\right)=3\sin \left(\mathrm{x}\right)-4{\sin }^3\mathrm{x}=\mathrm{sinx}\left(3-4{\sin }^2\mathrm{x}\right) \\ \cos \left(3\mathrm{x}\right)=4{\cos }^3\left(\mathrm{x}\right)-3\cos \left(\mathrm{x}\right)=\mathrm{cosx}\left(4{\cos }^2\mathrm{x}-3\right) \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{sinx}\left(3-4{\sin }^2\mathrm{x}\right).{\sin }^3\mathrm{x}+\mathrm{cosx}\left(4{\cos }^2\mathrm{x}-3\right).{\cos }^3\mathrm{x} \\ =\left(3-4{\sin }^2\mathrm{x}\right).{\sin }^4\mathrm{x}+\left(4{\cos }^2\mathrm{x}-3\right).{\cos }^4\mathrm{x} \\ =3{\sin }^4\mathrm{x}-3{\cos }^4\mathrm{x}-4{\sin }^6\mathrm{x}+4{\cos }^6\mathrm{x} \\ =3\left({\sin }^4\mathrm{x}-{\cos }^4\mathrm{x}\right)-4\left({\sin }^6\mathrm{x}-{\cos }^6\mathrm{x}\right) \\ =3\left\{{\left({\sin }^2\mathrm{x}\right)}^2-{\left({\cos }^2\mathrm{x}\right)}^2\right\}-4\left\{{\left({\sin }^2\mathrm{x}\right)}^3-{\left({\cos }^2\mathrm{x}\right)}^3\right\} \\ =3\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)-4\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)\left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}+{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}\right) \\ =3\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)-4\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)\left({\sin }^4\mathrm{x}+{\cos }^4\mathrm{x}+{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}\right) \left[{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1\right] \\ =3\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)-4\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)\left({\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)}^2-2{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}\right) \\ =3\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)-4\left({\sin }^2\mathrm{x}-{\cos }^2\mathrm{x}\right)\left(1-{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}\right) \\ =4\left({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\right)\left(1-{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}\right) -3\left({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\right) \\ =\left({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\right)\left[4-4{\sin }^2\mathrm{x}.{\cos }^2\mathrm{x}-3\right] \\ =\cos \left(2\mathrm{x}\right)\left[1-4\left(1-{\cos }^2\mathrm{x}\right).{\cos }^2\mathrm{x}\right] \left[\mathrm{As} \left({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}\right)=\cos \left(2\mathrm{x}\right)\right] \\ =\cos \left(2\mathrm{x}\right)\left[1-4{\cos }^2\mathrm{x}+4{\cos }^4\mathrm{x}\right] \\ =\cos \left(2\mathrm{x}\right)\left[{\left(2{\cos }^2\mathrm{x}\right)}^2+1^2-2\times \left(2{\cos }^2\mathrm{x}\right)\times 1\right] \\ =\cos \left(2\mathrm{x}\right){\left[2{\cos }^2\mathrm{x}-1\right]}^2 \\ =\cos \left(2\mathrm{x}\right){\left[\cos \left(2\mathrm{x}\right)\right]}^2 \left[2{\cos }^2\mathrm{x}-1=\cos \left(2\mathrm{x}\right)\right] \\ ={\cos }^3\left(2\mathrm{x}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

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