pouzdrotf

2022-06-30

Prove ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{4}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$

Shawn Castaneda

Expert

As ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$ we have
${\mathrm{cos}}^{4}\theta -{\mathrm{sin}}^{4}\theta =\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta$
That is,
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{4}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$

Araceli Clay

Expert

I took the long-haul approach for you since it's nice and clear to see. There is a lot of play around with the fact: ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$ rearranged into ${\mathrm{sin}}^{2}\theta =1-{\mathrm{cos}}^{2}\theta$ and ${\mathrm{cos}}^{2}\theta =1-{\mathrm{sin}}^{2}\theta$
We can see that: ${\mathrm{cos}}^{4}\theta ={\mathrm{cos}}^{2}\theta {\mathrm{cos}}^{2}\theta =\left(1-{\mathrm{sin}}^{2}\theta \right)\left(1-{\mathrm{sin}}^{2}\theta \right)=1-2{\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{4}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{2}\theta +\left(1-2{\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta \right)={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{2}\theta +1-2{\mathrm{sin}}^{2}\theta +{\mathrm{sin}}^{4}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{4}\theta -{\mathrm{sin}}^{2}\theta +1={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{4}\theta -\left(1-{\mathrm{cos}}^{2}\theta \right)+1={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{4}\theta -1+{\mathrm{cos}}^{2}\theta +1={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$
${\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{2}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{4}\theta$

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