Expert answers to "Prove that by setting these two expressions equal..."

College
Algebra
Linear algebra
Alternate coordinate systems

jernplate8

Answered question

2020-11-09

Prove that by setting these two expressions equal to another, the result is an identity:

d = \cos^{- 1} (\sin (L T_{1}) \sin (L T_{2}) + \cos (L T_{1}) \cos (L T_{2}) \cos (\ln_{1} - \ln_{2}))

d = 2 \sin^{- 1} (\sqrt{\sin^{2} (\frac{L T_{1} - L T_{2}}{2}) + \cos (L T_{1}) \cos (L T_{2}) \sin^{2} (\frac{\ln_{1} - \ln_{2}}{2})})

Answer & Explanation

curwyrm

Skilled2020-11-10Added 87 answers

Concept used:
$\cos θ = 1 - 2 \sin^{2} (\frac{θ}{2})$
$\cos (A - B) = \cos B \cos A + \sin B \sin A$
Proof:
$d = 2 \sin^{- 1} (\sqrt{\sin^{2} (\frac{L T_{1} - L T_{2}}{2}) + \cos (L T_{1}) \cos (L T_{2}) \sin^{2} (\frac{\ln_{1} - \ln_{2}}{2})})$
$\Rightarrow \frac{\sin d}{2} = (\sqrt{\sin^{2} (\frac{L T_{1} - L T_{2}}{2}) + \cos (L T_{1}) \cos (L T_{2}) \sin^{2} (\frac{\ln_{1} - \ln_{2}}{2})})$
As,
$\cos θ = 1 - 2 \sin^{2} (\frac{θ}{2})$
So,
$(\sin (L T_{1}) \sin (L T_{2}) + \cos (L T_{1}) \cos (L T_{2}) \cos (\ln_{1} - \ln_{2}))$
$= 1 - 2 {(\sqrt{\sin^{2} (\frac{L T_{1} - L T_{2}}{2}) + \cos (L T_{1}) \cos (L T_{2}) \sin^{2} (\frac{\ln_{1} - \ln_{2}}{2})})}^{2}$
Therefore,
$(\sin (L T_{1}) \sin (L T_{2}) + \cos (L T_{1}) \cos (L T_{2}) \cos (\ln_{1} - \ln_{2}))$
$= 1 - 2 (\sin^{2} (\frac{L T_{1} - L T_{2}}{2}) + \cos (L T_{1}) \cos (L T_{2})$

Do you have a similar question?Recalculate according to your conditions!

New Questions in Linear algebra

Didn't find what you were looking for?