Question

If $T_n={\sin }^{n}x+{\cos }^{n}x$, then find the value of $6T_{10}-15T_8+10T_6$.

Answer 1

$T_n={\sin }^{n}x+{\cos }^{n}x$

$T_{10}={\sin }^{10}x+{\cos }^{10}x$

$T_6={\sin }^{6}x+{\cos }^{6}x$

$T_8={\sin }^{8}x+{\cos }^{8}x$

$\Rightarrow 6T_{10}-15T_8+10T_6$

$T_6={\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3$

$\because a^3+b^3=(a+b)(a^2+b^2-ab)$

$T_6=({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)$

$T_6={\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x$

$\because a^2+b^2={(a+b)}^2-2ab$

$\Rightarrow T_6={\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2-{\sin }^{2}x{\cos }^{2}x$

$={({\sin }^{2}x+{\cos }^{2}x)}^2-3{\sin }^{2}x{\cos }^{2}x$

$=1-3{\sin }^{2}x{\cos }^{2}x$

similarly, we can calculate $T_8$ & $T_{10}$

$T_8=({\sin }^{6}x+{\cos }^{6}x)({\sin }^{2}x+{\cos }^{2}x)-{\sin }^{2}x{\cos }^{2}x({\sin }^{4}x+{\cos }^{4}x)$

$=1-3{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x(1-2{\sin }^{2}x{\cos }^{2}x)$

$=1-4{\sin }^{2}x{\cos }^{2}x+2{\sin }^{4}x{\cos }^{4}x$

$T_{10}=({\sin }^{6}x+{\cos }^{6}x)({\sin }^{4}x+{\cos }^{4}x)-{\sin }^{4}x{\cos }^{4}x({\sin }^{2}x+{\cos }^{2}x)$

$=(1-3{\sin }^{2}x{\cos }^{2}x)(1-2{\sin }^{2}x{\cos }^{2}x)-{\sin }^{4}x{\cos }^{4}x$

$=1-5{\sin }^{2}x{\cos }^{2}x+5{\sin }^{4}x{\cos }^{4}x$

putting $T_6,T_8$ and $T_{10}$ in $6T_{10}-15T_8+10T_6$ we get $1$.

Hence, the answer is $1$.