If + =m, then prove that ^6x+ ^6x= 4-3(m^2-1)^2 4 , where m^2 2

Question

If $\sin\text{x}+\sin\cos\text{x}=\text{m},$ then prove that $\sin^6\text{x}+\cos^6\text{x}=\frac{4-3(\text{m}^2-1)^2}{4},$ where $\text{m}^2\leq2$

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Answer

To show: $\sin^6\text{x}+\cos^6\text{x}=\frac{4-3(\text{m}^2-1)^2}{4},$ $\text{ where m}^2\leq2$
Since, $\sin\text{x}+\cos\text{x}=\text{m}\cdots(\text{i})$
$\Rightarrow(\sin\text{x}+\cos\text{x})^2=\text{m}^2$
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{m}^2$
$\Rightarrow1+2\sin\text{x}\cos\text{x}=\text{m}^2$ $(\because\sin^2\text{x}+\cos^2\text{x}=1)$
$\Rightarrow2\sin\text{x}\cos\text{x}=\text{m}^2-1$
$\Rightarrow\sin\text{x}\cos\text{x}=\frac{\text{m}^2-1}{2}\cdots(\text{ii})$
$\therefore\text{L.H.S}=\sin^6\text{x}+\cos^2\text{x}$
$=(\sin^2\text{x})^3+(\cos^3\text{x})^3$
$=(\sin^2\text{x}+\cos^2\text{x})(\sin^2\text{x})^2+(\cos^2\text{x})^2-\sin^2\text{x}\cos^2\text{x}$
$=1.((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ \ +2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x}-\sin^2\text{x}\cos^2\text{x})$ $(\text{adding and subtracting }2\sin^2\text{x}\cos^2\text{x})$
$=(\sin^2\text{x}+\cos^2\text{x})^2-3\sin^2\text{x}\cos^2\text{x}$
$=1-3\sin^2\text{x}\cos^2\text{x}$
$=1-3(\sin\text{x}\cos\text{x})^2$
$=1-3\frac{(\text{m}^2-1)^2}{4}$ $\text{(from (ii))}$
$=\frac{4-3(\text{m}^2-1)^2}{4}, \text{ where m}^2\leq2$
$=\text{R.H.S}$
$\text{Proved}$

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