Question

If $T_n=si{n}^n\theta +co{s}^n\theta$, prove that

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

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

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

Answer 1

(i)We have,

$T_n=si{n}^n\theta +co{s}^n\theta .....\left(i\right)$

To show: $\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_3}$

$LHS=\frac{T_3-T_5}{T_1}$

$=\frac{(si{n}^3\theta +co{s}^3\theta )-(si{n}^5\theta +co{s}^5\theta )}{sin\theta +cos\theta }$ [Substituting the values of $T_3,T_5$ and $T_1$ from (i)]

$=\frac{si{n}^3\theta -si{n}^5\theta +co{s}^3\theta -co{s}^5\theta }{sin\theta +cos\theta }=\frac{si{n}^3\theta (1-si{n}^2\theta )+co{s}^3\theta (1-co{s}^2\theta )}{sin\theta +cos\theta }$

$=\frac{si{n}^3\theta co{s}^2\theta +co{s}^3\theta si{n}^2\theta }{sin\theta +cos\theta }$ $\left[\begin{array}{c}\because 1-si{n}^2\theta =co{s}^2\theta \\ and1-co{s}^2\theta =si{n}^2\theta \end{array}\right]$

$=\frac{si{n}^2\theta co{s}^2\theta +(sin\theta +cos\theta )}{sin\theta +cos\theta }=si{n}^2\theta co{s}^2\theta$

$RHS=\frac{si{n}^5\theta +co{s}^5\theta -(si{n}^7\theta +co{s}^7\theta )}{si{n}^3\theta +co{s}^3\theta }=\frac{si{n}^5\theta -si{n}^7\theta +co{s}^5\theta -co{s}^7\theta }{si{n}^3\theta +co{s}^3\theta }$

$=\frac{si{n}^5\theta (1-si{n}^2\theta )+co{s}^5\theta (1-co{s}^2\theta )}{si{n}^3\theta +co{s}^3\theta }=\frac{si{n}^5\theta co{s}^2\theta +co{s}^5\theta si{n}^2\theta }{si{n}^3\theta +co{s}^3\theta }$

$=\frac{si{n}^2\theta co{s}^2\theta (si{n}^3\theta +co{s}^2\theta )}{si{n}^3\theta +co{s}^3\theta }=si{n}^2\theta co{s}^2\theta$

LHS =RHS Hence proved.

$\left(ii\right)LHS=2T_6-3T_4+1$

$=2(si{n}^6\theta +co{s}^6\theta )-3(si{n}^4\theta +co{s}^4\theta )+1$

$=2\left[\right(si{n}^2\theta {)}^3+(co{s}^2\theta {)}^3-3(si{n}^2\theta {)}^2+\left(co{s}^2\theta {)}^2\right]+1$

$=2\left[\right(si{n}^2\theta +co{s}^2\theta \left)\right(si{n}^2\theta {)}^2+(co{s}^2\theta {)}^2-(si{n}^2\theta co{s}^2\theta \left)\right]-3\left[\right(si{n}^2\theta {)}^2+(co{s}^2\theta {)}^2+2si{n}^2\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta ]+1$

[Using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and adding and subtracting $2si{n}^2\theta co{s}^2\theta ]$

$=2\left[\right(si{n}^2\theta +co{s}^2\theta {)}^2-3si{n}^2\theta co{s}^2\theta -3(1-2si{n}^2\theta co{s}^2\theta )+1]$

$=2(1-3si{n}^2\theta co{s}^2\theta )-3+6si{n}^2\theta co{s}^2\theta +1$

$=2-6si{n}^2\theta co{s}^2\theta -2+6si{n}^2\theta co{s}^2\theta$

=0

=RHS Hence proved.

$\left(iii\right)LHS=6T_{10}-15T_8+10T_6-1$

$=6(si{n}^{10}\theta +co{s}^{10}\theta )-15(si{n}^8\theta +co{s}^8\theta )+10(si{n}^6\theta +co{s}^6\theta )-1$

$=6si{n}^{10}\theta -15si{n}^8\theta +10si{n}^6\theta +6co{s}^{10}\theta -15co{s}^8\theta +10co{s}^6\theta -1$

$=si{n}^6\theta (6si{n}^4\theta -15si{n}^2\theta +10)+co{s}^6\theta (6co{s}^4\theta -15co{s}^2\theta +10)-(si{n}^2\theta +co{s}^2\theta {)}^3$

$[\because 1=si{n}^2\theta +co{s}^2\theta ]$

$=si{n}^6\theta (6si{n}^4\theta -15si{n}^2\theta +10)+co{s}^6\theta (6co{s}^4\theta -15co{s}^2\theta +10)-(si{n}^6\theta +co{s}^6\theta +3si{n}^2\theta co{s}^2\theta (si{n}^2\theta +co{s}^2\theta \left)\right)$

[Using $(a+b{)}^2=a^3+b^3+3ab(a+b\left)\right]$

$=si{n}^6\theta (6si{n}^4\theta -15co{s}^2\theta +10-1)+co{s}^6\theta (6co{s}^4\theta -15co{s}^2\theta +10-1)-3si{n}^2\theta co{s}^2\times 1$

$[\because co{s}^2\theta +si{n}^2=1]=si{n}^6\theta (6si{n}^4\theta -9si{n}^2\theta -6si{n}^2\theta +9)+co{s}^6\theta (6co{s}^4\theta -9co{s}^2\theta -6co{s}^2\theta +9)-3si{n}^2\theta co{s}^2\theta$

[On splitting the middle term]

$=si{n}^6\theta \left[3si{n}^2\theta \right(2si{n}^2\theta -3)-3(2si{n}^2\theta -3\left)\right]+co{s}^6\theta \left[3co{s}^2\theta \right(2co{s}^2\theta -3)-3(2co{s}^2\theta -3\left)\right]-3si{n}^2\theta co{s}^2\theta$

$=si{n}^6\theta (2si{n}^2\theta -3)(3si{n}^2\theta -3)+co{s}^6\theta (2co{s}^2\theta -3)(3co{s}^2\theta -3)-3si{n}^2\theta co{s}^2\theta$

$=si{n}^6\theta \times (-3)(2si{n}^2\theta -3)(1-si{n}^2\theta )+co{s}^6\theta \times (-3)(2co{s}^2\theta -3)(1-co{s}^2\theta )-3si{n}^2\theta co{s}^2\theta$

$=-3si{n}^6\theta (2si{n}^2\theta -3)co{s}^2\theta -3co{s}^6\theta (2co{s}^2\theta -3)si{n}^2\theta -3si{n}^2\theta co{s}^2\theta$

$=6si{n}^8\theta +co{s}^2\theta +6si{n}^6\theta co{s}^2\theta -6co{s}^8\theta si{n}^2\theta +9co{s}^6\theta +si{n}^2\theta -3si{n}^2\theta co{s}^2\theta$

$=-6si{n}^2\theta +co{s}^2\theta (si{n}^6\theta +co{s}^6\theta )+9si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )-3si{n}^2\theta co{s}^2\theta$

$=-6si{n}^2\theta co{s}^2\theta \left[\right(si{n}^2\theta {)}^3+\left(co{s}^2\theta {)}^3\right]+9si{n}^2\theta co{s}^2\theta \left[\right(si{n}^2\theta {)}^2+\left(co{s}^2\theta {)}^2\right]-3si{n}^2\theta co{s}^2\theta$

$=-6si{n}^2\theta co{s}^2\theta (si{n}^2\theta +co{s}^2\theta )+(si{n}^4\theta +co{s}^4\theta -si{n}^2\theta co{s}^2\theta )+9si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )-3si{n}^2\theta co{s}^2\theta$ [Using $a^3+b^3=(a+b)(a^2+b^2-ab)]$

$=-6si{n}^2\theta co{s}^2\theta (si{n}^4\theta co{s}^4\theta -si{n}^2\theta co{s}^2\theta )+9si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )-3si{n}^2\theta co{s}^2\theta (\because co{s}^2\theta +sin62\theta =1)$

$=-6si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )+6si{n}^4\theta co{s}^4\theta +9si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )-3si{n}^2\theta co{s}^2\theta$

$=3si{n}^2\theta co{s}^2\theta (si{n}^4\theta +co{s}^4\theta )+6si{n}^4\theta co{s}^4\theta -3si{n}^2\theta co{s}^2\theta$

$=3si{n}^2\theta co{s}^2\theta \left[\right(si{n}^2\theta {)}^2+(co{s}^2\theta {)}^2+2sin62\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta ]+6si{n}^4\theta co{s}^4\theta -3sin62\theta co{s}^2\theta$ (adding and subtracting $2si{n}^2\theta co{s}^2\theta )$

$=3si{n}^2\theta co{s}^2\theta \left(\right(si{n}^2\theta +co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta )+6si{n}^4\theta co{s}^4\theta -3si{n}^2\theta co{s}^2\theta$

$=3si{n}^2\theta co{s}^2\theta (1-2si{n}^2\theta co{s}^2\theta )+6si{n}^4\theta co{s}^4\theta -3si{n}^2\theta co{s}^2\theta$

$=3si{n}^2\theta co{s}^2\theta (1-2si{n}^2\theta co{s}^2\theta )+6si{n}^4\theta co{s}^4\theta -3si{n}^2\theta co{s}^2\theta$

$=3si{n}^2\theta co{s}^2\theta -6si{n}^4\theta co{s}^4\theta +6si{n}^4\theta -3sin62\theta co{s}^2\theta$

=0 =RHS Hence proved.