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Question

$$\displaystyle \sin 36^{\circ} \sin 72^{\circ} .\sin 108^{\circ} .\sin 144^{\circ}=\frac{k}{16}.$$ Find the value of $$k$$.


Solution

$$\sin { { 36 }^{ o }\sin { { 72 }^{ o }\sin { { 108 }^{ o }\sin { { 144 }^{ o } }  }  }  } \\ =\sin { { 36 }^{ o }\sin { { 72 }^{ o }\sin { \left( { 90 }^{ o }+{ 18 }^{ o } \right) \sin { \left( { 90 }^{ o }+{ 34 }^{ o } \right)  }  }  }  } =\sin { { 36 }^{ o }\sin { { 72 }^{ o }\cos { { 18 }^{ o }\cos { { 54 }^{ o } }  }  }  } $$
$$\displaystyle =\frac { 1 }{ 4 } \left( 2\sin { { 36 }^{ o } } \cos { { 54 }^{ o } }  \right) \left( 2\sin { { 72 }^{ o }\cos { { 18 }^{ o } }  }  \right) $$
$$\displaystyle \\ =\frac { 1 }{ 4 } \left( \cos { \left( { 36 }^{ o }-{ 54 }^{ o } \right)  } -\cos { \left( { 36 }^{ o }+{ 54 }^{ o } \right)  }  \right) \left( \cos { \left( { 72 }^{ o }-{ 18 }^{ o } \right)  } -\cos { \left( { 72 }^{ o }+{ 18 }^{ o } \right)  }  \right) $$
$$\displaystyle =\frac { 1 }{ 4 } \left( \cos { { 18 }^{ o } } -\cos { { 90 }^{ o } }  \right) \left( \cos { { 54 }^{ o } } -\cos { { 90 }^{ o } }  \right) $$
$$\displaystyle =\frac { 1 }{ 4 } \cos { { 18 }^{ o }\cos { { 54 }^{ o }=\frac { 1 }{ 4 } \left( \sqrt { \frac { 5 }{ 8 } +\frac { \sqrt { 5 }  }{ 8 }  }  \right) \left( \sqrt { \frac { 5 }{ 8 } -\frac { \sqrt { 5 }  }{ 8 }  }  \right)  } =\frac { 5 }{ 16 }  } $$

Mathematics

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