jernplate8

2020-11-09

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

curwyrm

Concept used:
$\mathrm{cos}\theta =1-2{\mathrm{sin}}^{2}\left(\frac{\theta }{2}\right)$
$\mathrm{cos}\left(A-B\right)=\mathrm{cos}B\mathrm{cos}A+\mathrm{sin}B\mathrm{sin}A$
Proof:
$d=2{\mathrm{sin}}^{-1}\left(\sqrt{{\mathrm{sin}}^{2}\left(\frac{L{T}_{1}-L{T}_{2}}{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right){\mathrm{sin}}^{2}\left(\frac{{\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}}{2}\right)}\right)$
$⇒\frac{\mathrm{sin}d}{2}=\left(\sqrt{{\mathrm{sin}}^{2}\left(\frac{L{T}_{1}-L{T}_{2}}{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right){\mathrm{sin}}^{2}\left(\frac{{\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}}{2}\right)}\right)$
As,
$\mathrm{cos}\theta =1-2{\mathrm{sin}}^{2}\left(\frac{\theta }{2}\right)$
So,
$\left(\mathrm{sin}\left(L{T}_{1}\right)\mathrm{sin}\left(L{T}_{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right)\mathrm{cos}\left({\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}\right)\right)$
$=1-2{\left(\sqrt{{\mathrm{sin}}^{2}\left(\frac{L{T}_{1}-L{T}_{2}}{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right){\mathrm{sin}}^{2}\left(\frac{{\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}}{2}\right)}\right)}^{2}$
Therefore,
$\left(\mathrm{sin}\left(L{T}_{1}\right)\mathrm{sin}\left(L{T}_{2}\right)+\mathrm{cos}\left(L{T}_{1}\right)\mathrm{cos}\left(L{T}_{2}\right)\mathrm{cos}\left({\mathrm{ln}}_{1}-{\mathrm{ln}}_{2}\right)\right)$

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