Question

# Prove that by setting these two expressions equal to another, the result is an identity: displaystyle{d}={{cos}^{ -{{1}}}{left( sin{{left({L}{T}_{{1}}

Alternate coordinate systems
Prove that by setting these two expressions equal to another, the result is an identity:
$$\displaystyle{d}={{\cos}^{ -{{1}}}{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}} \cos{{\left({{\ln}_{{1}}-}{\ln}_{{2}}\right)}}\right)}}$$
$$\displaystyle{d}={2}{{\sin}^{ -{{1}}}{\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}}$$

2020-11-10

Concept used:
$$\displaystyle \cos{\theta}={1}-{2}{{\sin}^{2}{\left(\frac{\theta}{{2}}\right)}}$$
$$\displaystyle \cos{{\left({A}-{B}\right)}}= \cos{{B}} \cos{{A}}+ \sin{{B}} \sin{{A}}$$
Proof:
$$\displaystyle{d}={2}{{\sin}^{ -{{1}}}{\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}}$$
$$\displaystyle\Rightarrow\frac{ \sin{{d}}}{{2}}={\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}$$
As,
$$\displaystyle \cos{\theta}={1}-{2}{{\sin}^{2}{\left(\frac{\theta}{{2}}\right)}}$$
So,
$$\displaystyle{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}} \cos{{\left({{\ln}_{{1}}-}{\ln}_{{2}}\right)}}\right)}$$
$$\displaystyle={1}-{2}{\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}^{2}$$
Therefore,
$$\displaystyle{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}} \cos{{\left({{\ln}_{{1}}-}{\ln}_{{2}}\right)}}\right)}$$
$$\displaystyle={1}-{2}{\left({{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}\right)}$$
Now,
$$\displaystyle{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{\left\lbrace{1}-{2}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}\right\rbrace}\right)}$$
$$\displaystyle={1}-{2}{\left({{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}\right)}}\right.}$$
Thus,
$$\displaystyle{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}\right)}$$
$$\displaystyle={1}-{2}{\left({{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}\right)}$$
Hence,
$$\displaystyle \cos{{\left({L}{T}_{{1}}-{L}{T}_{{2}}\right)}}={1}-{2}{\left({{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}\right)}$$
which is an identity as $$\displaystyle \cos{\theta}={1}-{2}{{\sin}^{2}{\left(\frac{\theta}{{2}}\right)}}$$
Conclusion:
Hence, the result is an identity.