If + =a, then, 1. ^6x+ ^6x=............ 2. | - |=..............

Question

If $\sin\text{x}+\cos\text{x}=\text{a},$ then,

$\sin^6\text{x}+\cos^6\text{x}=............$
$|\sin\text{x}-\cos\text{x}|=..............$

✓

Answer

If $\sin\text{x}+\cos\text{x}=\text{a},$ then,

$\sin^6\text{x}+\cos^6\text{x}=$ $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$
$|\sin\text{x}-\cos\text{x}|=$ $\sqrt{2-\text{a}^2}.$

Solution:
Given that, $\sin\text{x}+\cos\text{x}=\text{a}$
squaring both sides, we get,
$(\sin\text{x}+\cos\text{x})^2=\text{a}^2$
$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow1+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow\sin\text{x}\cos\text{x}=\frac{\text{a}^2-1}{2}\dots(1)$

$\sin^6\text{x}+\cos^6\text{x}=(\sin^2\text{x})^3+(\cos^2\text{x})3$

$=(\sin^2\text{x}+\cos^2\text{x})^3-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})$
$=(1)^3-3\Big(\frac{\text{a}^2-1}{2}\Big)^2\cdot1=1-\frac{3(\text{a}^2-1)^2}{4}$
$=\frac{1}{4}[4-3(\text{a}^2-1)^2]$
Hence ,the value of the filler is $=\frac{1}{4}[4-3(\text{a}^2-1)^2]$

$|\sin\text{x}-\cos\text{x}|^2=\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}$

$=1-2\Big(\frac{\text{a}^2-1}{2}\Big)=1-(\text{a}^2-1)=1-\text{a}^2+1$
$=2-\text{a}^2$
$\therefore|\sin\text{x}-\cos\text{x}|=\sqrt{2-\text{a}^2}$ $[\therefore|\sin\text{x}-\cos\text{x}|>0]$
Hence , the value of the filler is $\sqrt{2-\text{a}^2}\cdot$

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