Question

Prove the following statements.
${\cos }^{6}A+{\sin }^{6}A=1-3{\sin }^{2}A+{\cos }^{2}A$.

Answer 1

We have,

LHS $={\sin }^{6}A+{\cos }^6$ A

$\Rightarrow$ LHS $=({\sin }^{2}A{)}^3+({\cos }^{2}A{)}^3$

using, $a^3+b^3=(a+b)(a^2-ab+b^2)$, we get

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

Using $(si{n}^{2}A+co{s}^{2}A=1)$, we get

$\Rightarrow$ LHS $=\left\{\right({\sin }^{2}A{)}^2+({\cos }^{2}A{)}^2+2{\sin }^{2}Aco{s}^{2}A-2{\sin }^{2}A{\cos }^{2}A-{\sin }^{}A{\cos }^{2}A\}$

$\Rightarrow$ LHS $=\left\{\right({\sin }^{2}A+{\cos }^{2}A{)}^2-3{\sin }^{2}A{\cos }^{2}A\}=1-3{\sin }^{2}A{\cos }^{2}A=RHS$

ALTERNATIVELY

LHS $=({\sin }^{2}A{)}^3+({\cos }^{2}A{)}^3$

$\Rightarrow$ LHS $=({\sin }^{2}A+{\cos }^{2}A)-3{\sin }^{2}A{\cos }^{2}A({\sin }^{2}A+{\cos }^{2}A)$ ...... $[\because a^3+b^3=(a+b{)}^3-3ab(a+b)]$

$\Rightarrow$ LHS $=1-3{\sin }^{2}A{\cos }^{2}A=$ RHS.