The value of 2 sin(12^{circ}) - sin(72^{circ} | vedclass

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(D) We have the expression $2 \sin(12^{\circ}) - \sin(72^{\circ})$.Using the identity $\sin(C) - \sin(D) = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(

Vedclass · Accepted Answer

$\frac{\sqrt{3}(1-\sqrt{5})}{4}$

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(D) We have the expression $2 \sin(12^{\circ}) - \sin(72^{\circ})$.Using the identity $\sin(C) - \sin(D) = 2 \cos\left(\frac{C+D}{2}ight) \sin\left(\frac{C-D}{2}ight)$,we rewrite the expression as:$\sin(12^{\circ}) + (\sin(12^{\circ}) - \sin(72^{\circ}))$$= \sin(12^{\circ}) - 2 \cos\left(\frac{12^{\circ}+72^{\circ}}{2}ight) \sin\left(\frac{72^{\circ}-12^{\circ}}{2}ight)$$= \sin(12^{\circ}) - 2 \cos(42^{\circ}) \sin(30^{\circ})$Since $\sin(30^{\circ}) = \frac{1}{2}$,we get:$= \sin(12^{\circ}) - \cos(42^{\circ})$$= \sin(12^{\circ}) - \sin(90^{\circ} - 42^{\circ}) = \sin(12^{\circ}) - \sin(48^{\circ})$Using $\sin(C) - \sin(D) = 2 \cos\left(\frac{C+D}{2}ight) \sin\left(\frac{C-D}{2}ight)$ again:$= 2 \cos\left(\frac{12^{\circ}+48^{\circ}}{2}ight) \sin\left(\frac{12^{\circ}-48^{\circ}}{2}ight)$$= 2 \cos(30^{\circ}) \sin(-18^{\circ}) = -2 \cos(30^{\circ}) \sin(18^{\circ})$Substituting $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(18^{\circ}) = \frac{\sqrt{5}-1}{4}$:$= -2 	imes \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{5}-1}{4}$$= -\frac{\sqrt{3}(\sqrt{5}-1)}{4} = \frac{\sqrt{3}(1-\sqrt{5})}{4}$

The value of $2 \sin(12^{\circ}) - \sin(72^{\circ})$ is

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