Question

If $T_n\theta +co{s}^n\theta$, prove that
i) $2T_6-3T_4+1=0$

Answer 1

$2T_b-3T_4+1=0$
LHS $2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1$
We know
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$\therefore si{n}^6\theta +co{s}^6\theta =(si{n}^2\theta {)}^3+(co{s}^2\theta {)}^3$
$=(si{n}^2\theta +co{s}^2\theta )(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta co{s}^2\theta )$
$=(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta .co{s}^2\theta )$
$\therefore 2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+12(si{n}^4\theta +co{s}^4\theta )-2si{n}^2\theta .co{s}^2\theta -3(si{n}^4\theta +co{s}^4\theta )+1$
$=1-(si{n}^4\theta +co{s}^4\theta )-2si{n}^2\theta co{s}^2\theta$
$=1-(si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta )$
$=1-(si{n}^2\theta +co{s}^2\theta {)}^2$
$=1-1=0=$RHS