Question

If $sin\theta +cos\theta =m$, then prove that

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

Answer 1

To show: $si{n}^6\theta +co{s}^6\theta =\frac{4-3(m^2-1{)}^2}{4}$, where $m^2\le 2$

Since, $sin\theta +cos\theta -m.....\left(i\right)$

$\Rightarrow (sin\theta +cos\theta {)}^2=m^2$

$\Rightarrow si{n}^2\theta +co{s}^2\theta +2sin\theta cos\theta =m^2$

$\Rightarrow 1+2sin\theta cos\theta =m^2(\because si{n}^2\theta +co{s}^2\theta =1)$

$\Rightarrow 2sin\theta cos\theta =m^2-1$

$\Rightarrow sin\theta cos\theta =\frac{m^2-1}{2}....\left(ii\right)$

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

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

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

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

[adding and subtracting $2si{n}^2\theta co{s}^2\theta ]$

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

$=1-3si{n}^2\theta co{s}^2\theta$

$=1-3(sin\theta cos\theta {)}^2$

$=1-3\frac{(m^2-1{)}^2}{4}$ [from (ii)]

$=\frac{4-3(m^2-1{)}^2}{4}$, where $m^2\le 2$

=RHS Hence proved.