मराठी

Write the Value of Sin π 15 Sin 4 π 15 Sin 3 π 10 - Mathematics

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प्रश्न

Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]

बेरीज
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उत्तर

\[\frac{\pi}{15} = 12^\circ, \frac{4\pi}{15} = 48^\circ, \frac{3\pi}{10} = 54^\circ\]
\[\sin12^\circ \sin48^\circ \sin54^\circ\]
\[ = \frac{1}{2}\left[ 2\sin12^\circ \sin48^\circ \right] \sin54^\circ\]
\[ = \frac{1}{2}\left[ \cos\left( 12^\circ - 48^\circ \right) - \cos\left( 12^\circ + 48^\circ \right) \right] \sin54^\circ\]
\[ = \frac{1}{2}\left[ \cos\left( - 36^\circ \right) - \cos60^\circ \right] \sin54^\circ\]
\[ = \frac{1}{2}\sin54^\circ\left[ \cos36^\circ - \frac{1}{2} \right]\]
\[ = \frac{1}{2}\left[ \sin\left( 90^\circ - 36^\circ \right) \cos36^\circ \right] - \frac{1}{4}\sin\left( 90^\circ - 36^\circ \right)\]
\[ = \frac{1}{2} \cos^2 36^\circ - \frac{1}{4}\cos36^\circ\]
\[ = \frac{1}{2} \left( \frac{\sqrt{5} + 1}{4} \right)^2 - \left( \frac{\sqrt{5} + 1}{16} \right) \left[ \cos36^\circ = \frac{\sqrt{5} + 1}{4} \right]\]
\[ = \frac{1}{2}\left( \frac{5 + 1 + 2\sqrt{5}}{16} \right) - \left( \frac{\sqrt{5} + 1}{16} \right)\]
\[ = \frac{6 + 2\sqrt{5}}{32} - \frac{\sqrt{5} + 1}{16}\]
\[ = \frac{6 + 2\sqrt{5} - 2\sqrt{5} - 2}{32}\]
\[ = \frac{4}{32}\]
\[ = \frac{1}{8}\]

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Transformation Formulae
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पाठ 8: Transformation formulae - Exercise 8.3 [पृष्ठ २०]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 8 Transformation formulae
Exercise 8.3 | Q 7 | पृष्ठ २०

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