Wronsonia8g

2022-07-09

In a triangle prove that ${\mathrm{sin}}^{2}\left(\frac{A}{2}\right)+{\mathrm{sin}}^{2}\left(\frac{B}{2}\right)+{\mathrm{sin}}^{2}\left(\frac{C}{2}\right)+2\mathrm{sin}\left(\frac{A}{2}\right)\mathrm{sin}\left(\frac{B}{2}\right)\mathrm{sin}\left(\frac{C}{2}\right)=1$

diamondogsaz

Expert

Let $x=\frac{A}{2}$, $y=\frac{B}{2}$ , $z=\frac{C}{2}$ , $x+y+z=\pi /2$
${\mathrm{sin}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(y\right)+{\mathrm{sin}}^{2}\left(z\right)+2\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)\mathrm{sin}\left(z\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(y\right)+{\mathrm{sin}}^{2}\left(\pi /2-x-y\right)+2\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)\mathrm{sin}\left(\pi /2-x-y\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(y\right)+{\mathrm{cos}}^{2}\left(x+y\right)+2\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(y\right)+{\mathrm{cos}}^{2}\left(x+y\right)+\left(\mathrm{cos}\left(x-y\right)-\mathrm{cos}\left(x+y\right)\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(y\right)+\mathrm{cos}\left(x-y\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}=\frac{1-\mathrm{cos}\left(2x\right)}{2}+\frac{1-\mathrm{cos}\left(2y\right)}{2}+\mathrm{cos}\left(x-y\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}=1-\frac{\mathrm{cos}\left(2x\right)+\mathrm{cos}\left(2y\right)}{2}+\mathrm{cos}\left(x-y\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{cos}\left(x-y\right)\mathrm{cos}\left(x+y\right)+\mathrm{cos}\left(x-y\right)\mathrm{cos}\left(x+y\right)\phantom{\rule{0ex}{0ex}}=1$

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