Question

If $T_n={\sin }^{n}x+{\cos }^{n}x$, then find the value of $2T_6-3T_4+10.$

Answer 1

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

$T_4={\sin }^{4}x+{\cos }^{4}x$

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

${2T}_6-3T_4+10=2({\sin }^{6}x+{\cos }^{6}x)-3({\sin }^{4}x+{\cos }^{4}x)+10$

$=2({\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3)-3({\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2)+10$

Formula $\left(i\right),\left(ii\right)$ and $\left(iii\right)$ will be used in the further calculations,

$\left(i\right)a^3+b^3=(a+b)(a^2-ab+b^2)$

$\left(ii\right)a^2+b^2={(a+b)}^2-2ab$

$\left(iii\right){\sin }^2\theta +{\cos }^2\theta =1$

Using these

$=2\left[({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)\right]-3{\sin }^{4}x-3{\cos }^{4}x+10$

$=2{\sin }^{4}x+2{\cos }^{4}x-2{\sin }^{2}x{\cos }^{2}x-3{\sin }^{4}x-3{\cos }^{4}x+10$

$=-{\sin }^{4}x-{\cos }^{4}x-2{\sin }^{2}x{\cos }^{2}x+10$

$=10-({\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2+2{\sin }^{2}x{\cos }^{2}x)$

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

$=10-1$

$=9$