How to find the smallest value of $\sin^4(\alpha)+\cos^4(\alpha)$?

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In my book only the exact $4$ values of trig functions are used when evaluating them, so after solving the above expression in the following way: $$\sin^4(\alpha)+\cos^4(\alpha)=(\sin^2(\alpha)+\cos^2(\alpha))^2-2\sin^2(\alpha)\cos^2(\alpha)=1-\sin^2(2\alpha),$$ I wrote let $\alpha$ equal $90^\circ$ $(\alpha=90^\circ),$ therefore $$1-\sin^2(2\alpha)=1-\sin^2(180^\circ)=1-0=1,$$ afterwards $\alpha=60^\circ,$ $$1-\sin^2(2\alpha)=1-\sin^2(120^\circ)=1-\frac{\sqrt{3}}{2}\approx 0.866,$$ when $\alpha=45^\circ,$ $$1-\sin^2(2\alpha)=1-\sin^2(90^\circ)=1-1=0,$$ when $\alpha=30^\circ,$ it gives the value of that of $1-\sin^2(120^\circ),$ so I concluded that the minimum value of the expression is $0,$ whereas

according to the book it is $\frac{1}{2}.$

I also thought of using the angle $15^\circ,$ which produces $\frac{1}{2},$ but then the other angles (such as $5^\circ$ etc) will have to be used which makes it ambiguous. I have no idea of how to get that solution. Any hints are welcome!

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Because actually $$\sin^4(\alpha)+\cos^4(\alpha)=(\sin^2(\alpha)+\cos^2(\alpha))^2-2\sin^2(\alpha)\cos^2(\alpha)=1-\frac12\sin^2(2\alpha),$$

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$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{(\sin2x)^2}2$$

Now $0\le\sin^22x\le1$

Alternatively using $\cos2x=1-2\sin^2x=2\cos^2x-1$,

$$(2\sin^2x)^2+(2\cos^2x)^2=(1-\cos2x)^2+(1+\cos2x)^2=2(1+\cos^22x)=2+1+\cos4x$$

Now $-1\le\cos4x\le1$

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$sin^2 (2\alpha)=4sin^2 (\alpha )cos^2(\alpha)$, so you need a factor of $\frac12$ to balance your equation. .. To minimize $1-\frac 12sin^2(2\alpha)$ we just need to maximize $\frac 12sin^2 (2\alpha ) $. As you know this occurs when $\alpha=45^\circ $ , and we get $1-\frac12sin^2 (2\cdot45^\circ)=1-\frac12 (1)=\frac12 $.

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By C-S $$\sin^4\alpha+\cos^4\alpha=\frac{1}{2}(1+1)(\sin^4\alpha+\cos^4\alpha)\geq\frac{1}{2}(\sin^2\alpha+\cos^2\alpha)^2=\frac{1}{2}.$$ The equality occurs for $x=45^{\circ}$, which says that $\frac{1}{2}$ is a minimal value.