Question

If $T_n=si{n}^n\theta +co{s}^n\theta$, prove that
$\left(i\right)\frac{T_3-T_5}{T_1}\left(ii\right)2T_6-3T_4+1=0$

Answer 1

(i) $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}LHS\frac{si{n}^3\theta +co{s}^3\theta -si{n}^5\theta -co{s}^5\theta }{sin\theta +cos\theta }=\frac{si{n}^3\theta -si{n}^5\theta +co{s}^3\theta -co{s}^5\theta }{sin\theta +cos\theta }=\frac{si{n}^3\theta (1-si{n}^2\theta )+co{s}^3\theta (1-co{s}^2\theta )}{sin\theta +cos\theta }=\frac{si{n}^3\theta .co{s}^2\theta +co{s}^3\theta .si{n}^2\theta }{sin\theta +cos\theta }$
$=\frac{si{n}^2\theta .co{s}^2\theta (sin\theta +cos\theta )}{(sin\theta +cos\theta )}=si{n}^2\theta .co{s}^2\theta RHS\frac{si{n}^5\theta +co{s}^5\theta -si{n}^7\theta .-co{s}^7\theta }{si{n}^3\theta +co{s}^3\theta }=\frac{si{n}^5\theta (1-si{n}^2\theta )+co{s}^5\theta (1-co{s}^2\theta )}{si{n}^3\theta +co{s}^3\theta }=\frac{si{n}^5\theta .co{s}^2\theta +co{s}^5\theta .si{n}^2\theta }{(si{n}^3\theta +co{s}^3\theta )}$
$=(si{n}^2\theta .co{s}^2\theta )=LHS$ .(Now take out commons terms)
(ii) $2T_b-3T_4+1=0$
LHS $2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1$
We know,
$a^3+b^3=(a+b)(a^2-ab+b^2)\therefore si{n}^6\theta +co{s}^6\theta =(si{n}^2\theta {)}^3+(co{s}^2\theta )=(si{n}^2\theta +co{s}^2\theta \left)\right(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta co{s}^2\theta )=(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta .co{s}^2\theta )$
$\therefore 2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1=2(si{n}^4\theta +co{s}^4\theta )-2si{n}^2\theta .co{s}^2\theta -3(si{n}^4\theta +co{s}^4\theta )+1=1-(si{n}^4\theta +co{s}^4\theta )-2si{n}^2\theta co{s}^2\theta =1-(si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta )=1-(si{n}^2\theta +co{s}^2\theta {)}^2=1-1=0=RHS$