Question

Prove that: $\sin 3A.{\sin }^{3}A+\cos 3A.{\cos }^{3}A={\cos }^{3}2A$

Answer 1

Recall

$\cos 3A=4{\cos }^{3}A-3cosA$

and

$\sin 3A=3sinA-4{\sin }^{3}A$

Using these in LHS,

$\therefore$ The L.H.S $={\cos }^{3}A(4{\cos }^{3}A-3cosA)+{\sin }^{3}A(3sinA-4{\sin }^{3}A)$

$\Rightarrow 4{\cos }^{6}A-3{\cos }^{4}A+3{\sin }^{4}A-4{\sin }^{6}A$

$\Rightarrow 4({\cos }^{6}A-{\sin }^{6}A)-3({\cos }^{4}A-{\sin }^{4}A)$

$\Rightarrow 4\{{\left({\cos }^{2}A\right)}^3-{\left({\sin }^{2}A\right)}^3\}-3\{{\left({\cos }^{2}A\right)}^2-{\left({\sin }^{2}A\right)}^2\}$

$\Rightarrow 4\{{\cos }^{2}A-{\sin }^{2}A\}\{{\left({\cos }^{2}A\right)}^2+{\cos }^{2}A{\sin }^{2}A+{\left({\sin }^{2}A\right)}^2\}-3({\cos }^{2}A-{\sin }^{2}A)({\cos }^{2}A+{\sin }^{2}A)$

$\Rightarrow ({\cos }^{2}A-{\sin }^{2}A)[4\{{\left({\cos }^{2}A\right)}^2+{\cos }^{2}A{\sin }^{2}A+{\left({\sin }^{2}A\right)}^2\}-3.1]$

$\Rightarrow ({\cos }^{2}A-{\sin }^{2}A)[{4\cos }^{2}A+{4\cos }^{2}A{\sin }^{2}A+{4\sin }^{4}A-3{({\cos }^{2}A+{\sin }^{2}A)}^2]$

$\Rightarrow ({\cos }^{2}A-{\sin }^{2}A)[{4\cos }^{4}A+{4\cos }^{2}A{\sin }^{2}A+{4\sin }^{4}A-3({\cos }^{4}A+2{\cos }^{2}A{\sin }^{2}A+{\sin }^{4}A)]$

$\Rightarrow ({\cos }^{2}A-{\sin }^{2}A)[{\cos }^{4}A-2{\cos }^{2}A{\sin }^{2}A+{\sin }^{4}A]$

$\Rightarrow ({\cos }^{2}A-{\sin }^{2}A){({\cos }^{2}A-{\sin }^{2}A)}^2$,

$\Rightarrow {({\cos }^{2}A-{\sin }^{2}A)}^3$

$\Rightarrow {\left(\cos 2A\right)}^3$

$\Rightarrow {\cos }^{3}2A$

$=RHS$ Hence proved