Prove that 1 2 4 sin 2 ⁡ ( 36 ∘ ) − 1 = cos ⁡ ( 72 ∘ )

Question

Prove that             1      2            4          sin      2        ⁡    (          36              ∘              )    −    1    =  cos  ⁡  (      72          ∘        )

Syllingbs · Accepted Answer

Note that we have the system of equations

{\begin{cases} \cos 72^{\circ} & = 2 \cos^{2} 36^{\circ} - 1, \\ \cos 36^{\circ} & = 1 - 2 \cos^{2} 72^{\circ} . \end{cases}

Adding the two equations together yields

\cos 72^{\circ} + \cos 36^{\circ} = 2 (\cos^{2} 36^{\circ} - \cos^{2} 72^{\circ}) ⟹ \cos 36^{\circ} - \cos 72^{\circ} = \frac{1}{2} .

Now subtracting the first equation from the second gives

\cos 36^{\circ} - \cos 72^{\circ} = 2 - 2 (\cos^{2} 36^{\circ} + \cos^{2} 72^{\circ}) ⟹ \cos^{2} 36^{\circ} + \cos^{2} 72^{\circ} = \frac{3}{4} .

To finish, note that this equality rearranges to

1 - \cos^{2} 36^{\circ} - \cos^{2} 72^{\circ} = \sin^{2} 36^{\circ} - \cos^{2} 72^{\circ} = \frac{1}{4} .

Hence

4 \sin^{2} 36^{\circ} - 1 = 4 \cos^{2} 72^{\circ}

, and taking the square root of both sides and dividing by 2 gives the desired result.

Scolfaro2y · Accepted Answer

We need  4      sin    2    ⁡      36    ∘    −  4      cos    2    ⁡      72    ∘    =  1Using   cos  ⁡  2  A  =  1  −  2      sin    2    ⁡  A  =  2      cos    2    ⁡  A  −  1  ,    ⟺    2  (  1  −  cos  ⁡      72    ∘    )  −  2  (  1  +  cos  ⁡      144    ∘    )  =  1As   cos  ⁡  (      180    ∘    −  B  )  =  −  cos  ⁡  B  ,    ⟺    cos  ⁡      36    ∘    −  cos  ⁡      72    ∘    =            1      2      Utilize Proving trigonometric equation   cos  ⁡  (      36    ∘    )  −  cos  ⁡  (      72    ∘    )  =  1      /    2

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