Question

If $T_n={\sin }^n\mathrm{\theta }+{\cos }^n\mathrm{\theta }$, prove that

(i) $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}$

(ii) $2T_6-3T_4+1=0$

(iii) $6T_{10}-15T_8+10T_6-1=0$

Answer 1

(i) LHS:

$$\begin{gathered} \frac{T_3-T_5}{T_1}=\frac{\left({\sin }^3\theta +{\cos }^3\theta \right)-\left({\sin }^5\theta +{\cos }^5\theta \right)}{\sin \theta +\cos \theta } \\ =\frac{{\sin }^3\theta -{\sin }^5\theta +{\cos }^3\theta -{\cos }^5\theta }{\sin \theta +\cos \theta } \\ =\frac{{\sin }^3\theta \left(1-{\sin }^2\theta \right)+{\cos }^3\theta \left(1-{\cos }^2\theta \right)}{\sin \theta +\cos \theta } \\ =\frac{{\sin }^3\theta .{\cos }^2\theta +c{\mathrm{os}}^3\theta .{\sin }^2\theta }{\sin \theta +\cos \theta } \\ =\frac{{\sin }^2\theta .{\cos }^2\theta \left(\sin \theta +c\mathrm{os}\theta \right)}{\sin \theta +\cos \theta } \\ ={\sin }^2\theta .{\cos }^2\theta \end{gathered}$$

RHS:
$$\begin{gathered} \frac{T_5-T_7}{T_3} \\ =\frac{\left({\sin }^5\theta +{\cos }^5\theta \right)-\left({\sin }^7\theta +{\cos }^7\theta \right)}{{\sin }^3\theta +{\cos }^3\theta } \\ =\frac{{\sin }^5\theta -\mathrm{si}{n}^7\theta +{\cos }^5\theta -{\cos }^7\theta }{{\sin }^3\theta +{\cos }^3\theta } \\ =\frac{{\sin }^5\theta \left(1-{\sin }^2\theta \right)+{\cos }^5\theta \left(1-{\cos }^2\theta \right)}{{\sin }^3\theta +{\cos }^3\theta } \\ =\frac{{\sin }^5\theta {\cos }^2\theta +{\cos }^5\theta {\sin }^2\theta }{{\sin }^3\theta +{\cos }^3\theta } \\ ={\sin }^2\theta .{\cos }^2\theta \end{gathered}$$

LHS = RHS

Hence proved.

(ii) LHS:

$$\begin{gathered} 2T_6-3T_4+1 \\ 2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ 2\left(si{n}^2\theta +{\cos }^2\theta \right)\left({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ 2.1.\left({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ 2{\sin }^4\theta +2{\cos }^4\theta -2{\sin }^2\theta {\cos }^2\theta -3{\sin }^4\theta -3{\cos }^4\theta +1 \\ -\left({\sin }^4\theta +{\cos }^4\theta \right)-{\sin }^2\theta {\cos }^2\theta +1 \\ -({\sin }^2\theta +{\cos }^2\theta {)}^2+1 \\ -1+1 \\ 0 \end{gathered}$$

Hence proved.

(iii) LHS:

$$\begin{gathered} 6T_{10}-15T_8+10T_6-1 \\ 6\left({\sin }^{10}\theta +{\cos }^{10}\theta \right)-15\left({\sin }^8\theta +{\cos }^8\theta \right)+10\left({\sin }^6\theta +{\cos }^6\theta \right)-1 \end{gathered}$$