How do i prove this sin 2 ⁡ ( 20 ) sin ⁡ ( 140 ) sin ⁡ ( 80 ) = cos ⁡ ( 80 ) cos ⁡ ( 20 )

Question

How do i prove this                     sin        2            ⁡      (      20      )              sin      ⁡      (      140      )      sin      ⁡      (      80      )        =            cos      ⁡      (      80      )              cos      ⁡      (      20      )

Jaydon Roth · Accepted Answer

sin        2            ⁡      (      20      )              sin      ⁡      (      140      )      sin      ⁡      (      80      )        =                    sin        2            ⁡      (      20      )              sin      ⁡      (      40      )      sin      ⁡      (      80      )        =            2      sin      ⁡      (      10      )      cos      ⁡      (      10      )      sin      ⁡      (      20      )              2      sin      ⁡      (      20      )      cos      ⁡      (      20      )      sin      ⁡      (      80      )        =            cos      ⁡      (      80      )      sin      ⁡      (      80      )              cos      ⁡      (      20      )      sin      ⁡      (      80      )        =            cos      ⁡      (      80      )              cos      ⁡      (      20      )        sin  ⁡  140  =  sin  ⁡  40  cos  ⁡  10  =  sin  ⁡  80  sin  ⁡  10  =  cos  ⁡  80  sin  ⁡  20  =  2  sin  ⁡  10  cos  ⁡  10

perlejatyh8 · Accepted Answer

Using compound angle and double angle formula yields

\begin{aligned} \sin 140^{\circ} \sin 80^{\circ} \cos 80^{\circ} & = \frac{1}{2} \sin 140^{\circ} \sin 160^{\circ} \\ = \frac{1}{2} (\cos 20^{\circ} - \cos 300^{\circ}) \\ = \frac{1}{2} (\cos 20^{\circ} - \cos 60^{\circ}) \\ = \frac{1}{2} \sin 20^{\circ} \sin 40^{\circ} \\ = \sin^{2} 20^{\circ} \cos 20^{\circ} \end{aligned}

Rearranging yields

\frac{\sin^{2} 20^{\circ}}{\sin 140^{\circ} \sin 80^{\circ}} = \frac{\cos 80^{\circ}}{\cos 20^{\circ}}

How do prove this (sin^2(20))/(sin(140)sin(80))=(cos(80))/(cos(20))

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