<s>[INST] Problem on Trigonometry If $ \sin x + \cos x = a$ , then find:
(i) $ \sin^6 x + \cos^6 x$
(ii) $ \lvert \sin x – \cos x \rvert $
I tried using the identity $a^3 + b^3$ for (i) but that didn't work and I ended up with an expression that was of no help. I have absolutely no idea on how to go about the problem. Any kind of help would be appreciated.
 [/INST] Given $\displaystyle \sin x+\cos x = a\Rightarrow (\sin x+\cos x)^2 = a^2\Rightarrow \sin x \cos x = \frac{a^2-1}{2}$
Using $(\sin ^2 x+\cos^2 x)^3 = (\sin^6 x+\cos^6 x)+3(\sin x\cos x)^2(\sin^2 x+\cos^2 x)$
So $\displaystyle 1=\sin^6 x+\cos^6 x+3\left(\frac{a^2-1}{2}\right)$
$\displaystyle \bullet\;  |\sin x-\cos x|^2 = (\sin x-\cos x)^2=1-2\sin x\cos x = 1-\frac{2(a^2-1)}{2}$
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