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 $\text{m}^2\leq2$

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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