<s>[INST] Range of $(\sin x)^6+ (\cos x)^6$ I need to find the range of the function $y = (\sin x)^6+ (\cos x)^6$
I did find the answer but working in a crude way rather than a methodical step by step approach. I give below the steps I used , please help with a methodical approach to such problems.
1) To find the max value of the function I noticed that in the range where x is $[0,2\pi]$ , when $\cos x$ hits $+1$ then $\sin x$ is $0$ , when $\cos x$ is $0$ 
 then $\sin x$ is $+1$ ..etc so the max value at any of these points could be either $1^6+0^6$ or $(-1)^6+0^6$ so the max value is 1.
to find the minimum I differentiated the function
$f' (x) = 6(\sin x)^5\cos x- 6(\cos x^5)\sin x =0$ , equating this to zero
we have $(\sin x)^4 = (\cos x)^4 => x = \pi/4 =>$ min value of function is $(\sin(\pi/4))^6 + (\cos(\pi/4))^6 = 1/4$ . so the range is $(\frac{1}{4},1)$.
Please can someone help with how can this type of problems be methodically approached ? - Thanks.
 [/INST] With $$a=\sin(x)$$ $$b=\cos(x)$$ we have $$a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)=(a^2+b^2)((a^2+b^2)^2-3a^2b^2)$$
Because of $a^2+b^2=1$ we have $$\sin^6(x)+\cos^6(x)=1-3\sin^2(x)\cos^2(x)=1-\frac{3}{4}\sin^2(2x)$$
Obviously the range of $\sin^2(2x)$ is $[0,1]$, so we get range $[\frac{1}{4},1]$
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