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Prove that $$\left(\sqrt{3} -4\sin\left(\frac{2\pi}{15}\right)\right)\cos\left(\frac{\pi}{30} \right) =\sin\left(\frac{\pi}{30} \right).$$

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$$\left(\sqrt3 -4\sin24^\circ\right)\cos6^\circ =\sin6^\circ$$ will be true

$$\iff\tan60^\circ-\tan6^\circ=4\sin24^\circ$$

$$\iff\sin\left(60^\circ-6^\circ\right)=4\sin24^\circ\cos60^\circ\cos6^\circ$$

$$\iff\sin54^\circ=2\sin24^\circ\cos6^\circ\text{ as }\cos60^\circ=\frac12$$

Using Werner Formulas, the Right Hand Side becomes $$\sin30^\circ+\sin18^\circ$$

So, we need to show $$\sin54^\circ-\sin18^\circ=\frac12\text{ as }\sin30^\circ=\frac12$$

**Method** $\#1:$

Multiplying $$S=\sin54^\circ-\sin18^\circ=\sin54^\circ+\sin\left(-18^\circ\right)$$ by $\displaystyle2\sin\left(\frac{54^\circ-(-18^\circ)}2\right)=2\sin36^\circ$

$$2\sin36^\circ\cdot S=2\sin36^\circ\cdot\sin54^\circ+2\sin36^\circ\cdot\sin\left(-18^\circ\right) $$

Applying Werner Formula in the Right Hand Side, $$2\sin36^\circ\cdot S=\cos18^\circ-\cos90^\circ+\cos54^\circ-\cos18^\circ$$

$$\implies S=\frac{\cos54^\circ}{2\sin36^\circ}=\frac12$$

**Method** $\#2:$

Using Prosthaphaeresis formula,

$$S=\sin54^\circ-\sin18^\circ=2\sin18^\circ\cos36^\circ$$

$$\implies S=\frac{2\sin18^\circ(\cos18^\circ)\cos36^\circ}{\cos18^\circ}$$

Using $\displaystyle\sin2A=2\sin A\cos A$

$$S=\frac{\sin36^\circ\cos36^\circ}{\cos18^\circ}=\frac{\sin72^\circ}{2\cos18^\circ}=\frac12 $$

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