2022-01-15

Given $\mathrm{cos}\left(a\right)+\mathrm{cos}\left(b\right)=1$, prove that $1-{s}^{2}-{t}^{2}-3{s}^{2}{t}^{2}=0$, where $s=\mathrm{tan}\left(\frac{a}{2}\right)$ and $t=\mathrm{tan}\left(\frac{b}{2}\right)$

Elaine Verrett

Expert

Alternatively:
$\mathrm{cos}a=2{\mathrm{cos}}^{2}\frac{a}{2}-1$
${\mathrm{cos}}^{2}\frac{a}{2}=\frac{1}{1+{\mathrm{tan}}^{2}\frac{a}{2}}=\frac{1}{1+{s}^{2}}$
$\mathrm{cos}a+\mathrm{cos}b=2{\mathrm{cos}}^{2}\frac{a}{2}+2{\mathrm{cos}}^{2}\frac{b}{2}-2=$
$\frac{2}{1+{s}^{2}}+\frac{2}{1+{t}^{2}}-2=1$

RizerMix

Expert

$\frac{1-{t}^{2}}{1+{t}^{2}}+\frac{1-{s}^{2}}{1+{s}^{2}}=1$ $1+{s}^{2}-{t}^{2}-{s}^{2}{t}^{2}+1+{t}^{2}-{s}^{2}-{s}^{2}{t}^{2}=1+{s}^{2}+{t}^{2}+{s}^{2}{t}^{2}$ $⇒1-{s}^{2}-{t}^{2}-3{s}^{2}{t}^{2}=0$

alenahelenash

Expert

Hint: $\frac{1-{s}^{2}}{1+{s}^{2}}+\frac{1-{t}^{2}}{1+{t}^{2}}=1$ $⇔\frac{1-{s}^{2}}{1+{s}^{2}}=1-\frac{1-{t}^{2}}{1+{t}^{2}}$ $⇔\left(1-{s}^{2}\right)\left(1+{t}^{2}\right)=2{t}^{2}\left(1+{s}^{2}\right)$