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
ANSWERED
asked 2020-11-09
Prove that by setting these two expressions equal to another, the result is an identity:
\(\displaystyle{d}={{\cos}^{ -{{1}}}{\left( \sin{{\left({L}{T}_{{1}}\right)}} \sin{{\left({L}{T}_{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}} \cos{{\left({{\ln}_{{1}}-}{\ln}_{{2}}\right)}}\right)}}\)
\(\displaystyle{d}={2}{{\sin}^{ -{{1}}}{\left(\sqrt{{{{\sin}^{2}{\left(\frac{{{L}{T}_{{1}}-{L}{T}_{{2}}}}{{2}}\right)}}+ \cos{{\left({L}{T}_{{1}}\right)}} \cos{{\left({L}{T}_{{2}}\right)}}{{\sin}^{2}{\left(\frac{{{{\ln}_{{1}}-}{\ln}_{{2}}}}{{2}}\right)}}}}\right)}}\)

Answers (1)

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.

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