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Find the minimum value of $\sin^4 \theta + \cos^4 \theta $.

My Attempt: I tried to get some ideas from here: mrunal.org/2013/07/trigonometry-finding-minimum-maximum-values-for-ssc-cgl-made-easy-without-differentiation.html but couldn't.

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  • $\begingroup$ Hint: double angle identity for co/sine.$$\cos^2x=\dfrac{1+\cos2x}2$$$$\sin^2x=\dfrac{1-\cos2x}2$$ $\endgroup$ – user170231 May 20 '17 at 15:11
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We have that $\sin^2\theta+\cos^2\theta=1$ also $$(\sin^2\theta+\cos^2\theta)^2=\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta=1$$ So we have that $$\sin^4\theta+\cos^4\theta=1-\frac{1}{2}\sin^2(2\theta)$$ The minimum value is attained when $\sin^2(2\theta)$ attains it maximum. In general what is the max of the $\sin$ function?

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If you let $u=\sin^2\theta$ then $0\le u\le 1$ and $$\sin^4\theta+\cos^4\theta=u^2+(1-u)^2=1-2u+2u^2$$ so all you have to do in minimise $1-2u+2u^2$ on the interval $[0,1]$.

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Switch to polar (disregard the origin), $\sin (\theta)=\frac{y}{r}$ and $\cos (\theta)=\frac{x}{r}$. Then use the Am-GM inequality,


$$f(x,y)=\cos^4 (\theta)+\sin^4 (\theta)=\frac{x^4+y^4}{r^4} \geq 2\sqrt{\frac{x^4}{r^4}\frac{y^4}{r^4}}$$

Equality happens if and only if,

$$\frac{x^4}{r^4}=\frac{y^4}{r^4}$$

Hence we get,

$$x=\pm y$$

And we may write,

$$\text{min} f(x,y)=2\sqrt{\frac{x^8}{(x^2+x^2)^4}}=\frac{2}{2^2}=\frac{1}{2}$$


A similar argument shows,

$$\cos^{2n} (\theta)+\sin^{2n} (\theta) \geq 2\sqrt{\frac{x^{4n}}{(x^{2}+x^{2})^{2n}}}=\frac{2}{2^n}$$

$$=2^{1-n}$$

For all integer $n$ such that $n \geq 1$.

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Letting $x=\cos^2(\theta)$, note that $\sin^4(\theta) = (1-x)^2$. Then we want to minimize $$x^2+(1-x)^2 = 2x^2-2x+1$$ For a quadratic with a positive leading coefficient, the minimum occurs at the vertex, i.e. when $$x = -\frac{b}{2a} = \frac{2}{4} = \frac 1 2$$ So, the original function is minimized when $\cos^2(\theta) = \frac 1 2$.

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