2022/036) Let $t_n= \sin^n \theta + \cos^n \theta$ show that $6t_{10} − 15t_8 + 10t_6 = 1$ .

Let us put $p = \sin ^2 \theta \cos^2 \theta$  

We have $t_2 = \sin ^2 \theta + \cos^2 \theta = 1 \cdots(1)$

and $t_4 = (\sin^4x + \cos^4 x) = (\sin ^2 x + \cos^2 x) - 2 \sin ^2 x +\cos^2 x = 1 - 2p$ 

Now we an sert 

$t_n t_2 = ( \sin ^n \theta + \cos^n \theta)(\sin ^2 \theta + \cos^2 \theta) = \sin  ^{(n+2)}\theta +  \cos^{(n+2)} + \sin ^n \theta \cos^2 \theta +    \sin ^2 \theta \cos^n \theta$

or $t_n = t_{(n+2)} + t_{(n-2)}p$

or $t_(n+2) = t_n - t(n-2)p$

Using this we have

$t_6 = t_4 - t_2p = t_4 - p = 1- 3p$

 $t_8 = t_6 - t_4p = 1-3p -  (1-2p ) p^2 = 1- 4p + 2p^2$

$t_{10} = t_8 - t_6p  = (1-4p +2p^2) - (1-3p)p = 1 - 5p + 5p^2$

so $6t_{10} − 15t_8 + 10t_6 = 6(1-5p + 5p^2 ) - 15(1- 4p + 2p^2 ) + 10(1-3p)  = 1$