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