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Transformation Formulae — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae1 Mark
Question
Write the value of $\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}.$
✓

Answer

$\sin12^\circ\sin48^\circ\sin54^\circ$ $=\ \frac{1}{2}(2\sin12^\circ\sin48^\circ)\sin54^\circ$ $=\ \frac{1}{2}[\cos(48^\circ-12^\circ)-\cos(48^\circ+12^\circ)]\sin54^\circ$ $\{\text{Since }2\sin\text{A}\cos\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})\}$ $=\ \frac{1}{2}[\cos36^\circ-\cos60^\circ]\sin54^\circ$ $=\ \frac{1}{2}\Big[\cos36^\circ\sin54^\circ-\frac{1}{2}\sin45^\circ\Big]$ $=\ \frac{1}{4}[2\cos36^\circ\sin54^\circ-\sin45^\circ]$ $=\ \frac{1}{4}[\sin(54^\circ+36^\circ)+\sin(54^\circ-36^\circ)-\sin54^\circ]$ $\{\text{Since }2\sin\text{A}\cos\text{B}=\sin(\text{A}+\text{B})+\sin(\text{A}-\text{B})\}$ $=\ \frac{1}{4}[\sin90^\circ+\sin18^\circ-\sin54^\circ]$ $=\ \frac{1}{4}[\sin90^\circ-\sin54^\circ+\sin18^\circ]$ $=\ \frac{1}{4}\Big[1-\frac{\sqrt5+1}{4}+\frac{\sqrt5-1}{4}\Big]$ $\Big\{\text{Since }\sin18^\circ=\frac{\sqrt5-1}{4},\sin54^\circ\frac{\sqrt5+1}{4}\Big\}$ $=\ \frac{1}{4}\Big[\frac{4-\sqrt5-1+\sqrt5-1}{4}\Big]$ $=\ \frac{1}{4}\Big[\frac{2}{4}\Big]$ $=\ \frac{1}{8}$ Thus, $\sin12^\circ\sin48^\circ\sin54^\circ=\frac{1}{8}$

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