Question

Prove that: ${\sin }^{6}x+{\cos }^{6}x=1-3{\sin }^2{\cos }^{2^{}}x$

Answer 1

Trigonometry identity:

Let us consider

LHS:${\sin }^{6}x+{\cos }^6={({\sin }^{2}x)}^3+{({\cos }^{2}x)}^3$

By using the formula,$\mathbf{}{\mathbf{a}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathbf{ab}\mathbf{)}$

$$\begin{gathered} =({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})[{({\sin }^2\mathrm{x})}^2+{({\cos }^2\mathrm{x})}^2–{\sin }^{2}x{\cos }^{2}x] \\ \mathrm{By}\mathrm{using}\mathrm{the}\mathrm{formula},{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1\mathrm{and}{\mathrm{a}}^2+{\mathrm{b}}^2={(\mathrm{a}+\mathrm{b})}^2–2\mathrm{ab} \\ =1\times [{({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x})}^2–2{\sin }^2\mathrm{x}{\cos }^2\mathrm{x}–{\sin }^2\mathrm{x}{\cos }^2\mathrm{x}] \\ =1^2–3{\sin }^2\mathrm{x}{\cos }^2\mathrm{x} \\ =1–3{\sin }^2\mathrm{x}{\cos }^2\mathrm{x} \end{gathered}$$

$=$ RHS

LHS = RHS

Hence it is proved that ${\sin }^{6}x+{\cos }^{6}x=1-3{\sin }^2{\cos }^{2^{}}x$.