Banguizb

2022-07-09

Prove
$\frac{4co{s}^{2}\left(2x\right)-4co{s}^{2}\left(x\right)+3si{n}^{2}\left(x\right)}{4co{s}^{2}\left(\frac{5\pi }{2}-x\right)-si{n}^{2}2\left(x-\pi \right)}=\frac{8cos\left(2x\right)+1}{2\left(cos\left(2x\right)-1\right)}$

thatuglygirlyu

Expert

In the numerator,
$4{\mathrm{cos}}^{2}\left(2x\right)-7{\mathrm{cos}}^{2}\left(x\right)+3$
As suggested in the comments, use,
${\mathrm{cos}}^{2}\left(x\right)=\frac{1+\mathrm{cos}\left(2x\right)}{2}$
Thus
$4{\mathrm{cos}}^{2}\left(2x\right)-7{\mathrm{cos}}^{2}\left(x\right)+3=4{\mathrm{cos}}^{2}\left(2x\right)-\frac{7}{2}-\frac{7}{2}\mathrm{cos}\left(2x\right)+3=\frac{1}{2}\left(8{\mathrm{cos}}^{2}\left(2x\right)-7\mathrm{cos}\left(2x\right)-1\right)=\frac{1}{2}\left(8\mathrm{cos}\left(2x\right)+1\right)\left(\mathrm{cos}\left(2x\right)-1\right)$
Dividing this with the expression for the denominator you have got gives the desired result.

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