Plot[((Sin[x])^3),{x,0,Pi},PlotStyle->{RGBColor[0,0,0.7],Thickness[0.01]},AspectRatio->Automatic,ImageSize->500]

From Wikipedia Power-reduction formula is

Plot[{((Sin[x])^3),(3/4)Sin[x]-(1/4)Sin[3 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.005]}},AspectRatio->Automatic,ImageSize->500]

Manipulate[Plot[{((Sin[x])^3),(3/4)Exp[-t]Sin[x]-(1/4)Exp[-((3^2))t]Sin[3 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.006]}},AspectRatio->Automatic,ImageSize->500],{t,0,5}]

((Integrate[((Sin[x])^3) Sin[x],{x,0,Pi}])/(Pi/2))

((Integrate[((Sin[x])^3) Sin[2x],{x,0,Pi}])/(Pi/2))

((Integrate[((Sin[x])^3) Sin[3x],{x,0,Pi}])/(Pi/2))

Table[((Integrate[((Sin[x])^3) Sin[k x],{x,0,Pi}])/(Pi/2)),{k,1,20}]

Table[((Integrate[((Sin[x])^4) Sin[k x],{x,0,Pi}])/(Pi/2)),{k,1,20}]

Table[((Integrate[((Sin[x])^5) Sin[k x],{x,0,Pi}])/(Pi/2)),{k,1,20}]

Plot[{((Sin[x])^5),(5/8)Sin[x]-(5/16)Sin[3 x]+(1/16)Sin[5 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.005]}},AspectRatio->Automatic,ImageSize->500]

Just to illustrate that even with two summands we get a good approximation for the blue function

Plot[{((Sin[x])^5),(5/8)Sin[x]-(5/16)Sin[3 x]+0(1/16)Sin[5 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.007]}},AspectRatio->Automatic,ImageSize->500]

Manipulate[Plot[{((Sin[x])^5),(5/8)Exp[-t]Sin[x]-(5/16)Exp[-((3^2))t]Sin[3 x]+(1/16)Exp[-((5^2))t]Sin[5 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.007]}},AspectRatio->Automatic,ImageSize->500],{t,0,5}]

For (((Sin[x]))^7) we have

Table[((Integrate[((Sin[x])^7) Sin[k x],{x,0,Pi}])/(Pi/2)),{k,1,20}]

Plot[{((Sin[x])^7),(35/64)Sin[x]-(21/64)Sin[3 x]+(7/64)Sin[5 x]-(1/64)Sin[7 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.007]}},AspectRatio->Automatic,ImageSize->500]

Just to illustrate that even with three summands we get a good approximation for the blue function

Plot[{((Sin[x])^7),(35/64)Sin[x]-(21/64)Sin[3 x]+(7/64)Sin[5 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.007]}},AspectRatio->Automatic,ImageSize->500]

Manipulate[Plot[{((Sin[x])^7),(35/64)Exp[-t]Sin[x]-Exp[-9 t](21/64)Sin[3 x]+(7/64)Exp[-25 t]Sin[5 x]-(1/64)Exp[-49 t]Sin[7 x]},{x,0,Pi},PlotStyle->{{RGBColor[0,0,0.7],Thickness[0.012]},{RGBColor[1,0,0],Thickness[0.007]}},AspectRatio->Automatic,ImageSize->500],{t,0,5}]

Does the solution satisfy the given PDE?

D[(35/64)Exp[-t]Sin[x]-Exp[-9 t](21/64)Sin[3 x]+(7/64)Exp[-25 t]Sin[5 x]-(1/64)Exp[-49 t]Sin[7 x],t]

D[(35/64)Exp[-t]Sin[x]-Exp[-9 t](21/64)Sin[3 x]+(7/64)Exp[-25 t]Sin[5 x]-(1/64)Exp[-49 t]Sin[7 x],{x,2}]

Are they equal?

D[(35/64)Exp[-t]Sin[x]-Exp[-9 t](21/64)Sin[3 x]+(7/64)Exp[-25 t]Sin[5 x]-(1/64)Exp[-49 t]Sin[7 x],t]==D[(35/64)Exp[-t]Sin[x]-Exp[-9 t](21/64)Sin[3 x]+(7/64)Exp[-25 t]Sin[5 x]-(1/64)Exp[-49 t]Sin[7 x],{x,2}]