<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] We have $$\sin^6 x +\cos^6 x =(\sin^2 x +\cos^2 x)(\sin^4 x + \cos^4 x -\sin^2 x \cos^2 x) $$ $$=(1)((\sin^2 x + \cos^2 x)^2-3\sin^2 x \cos^2 x) $$ $$=(1-3\sin^2 x\cos^2 x) \tag {1}$$
Now we know that $$(\sin x + \cos x)^2 =1 +2\sin x \cos x$$ Use this result to find the value of $(1)$.
For $(2) $, use the fact that $$(\sin x -\cos x)^2 =1-2\sin x\cos x$$ Hope it helps. 
</s>