# 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.

• Well, you can do some basic trig manipulation to view it as $\frac18(3\cos(4x)+5)$. This gives you a maximum of $\frac18(3+5)=1$ and a minimum of $\frac18(-3+5)=\frac14$. So the range is $[\frac14, 1]$. Jul 13, 2018 at 13:02

Write $s=\sin x,\,c=\cos x$ so $s^6+c^6=s^4-s^2c^2+c^4=1-3s^2c^2=1-\frac{3}{4}\sin^2 2x$, which has range $[\frac{1}{4},\,1]$.

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 $[\frac{1}{4},1]$

• Basically, this is J.G. 's answer. I made it somewhat more detailed. Jul 13, 2018 at 13:13
• Thank you , fully understood. Jul 13, 2018 at 16:41

The only thing wrong with your solution is that you didn't prove you actually found the maximum, but this is easy to fix. After you set the derivative equal to $0$ you didn't solve the resulting equation completely. You neglected the $x\text{'}$s that satisfy $\sin x\cos x=0.$ These are just the $x$ you need to prove your calculation of the maximum is correct.

To prove that the function actually assumes all the values between the minimum and the maximum, you need to appeal to the intermediate value theorem. Frankly, I can't remember if this is typically stated in a calculus course, or just taken for granted.

Use the substitution $$t=\sin^2x$$. So the problem changes into finding the range of $$f(t)=t^3+(1-t)^3=3t^2-3t+1$$ where $$0\le t\le1$$. The rest is easy.