Math, asked by mitrashreyasi9, 16 days ago

prove that Cosπ/17.Cos2π/17.cos4π/17.cos8π/17=1/16​

Answers

Answered by senboni123456
2

Step-by-step explanation:

We have,

 \sf{cos \left(  \dfrac{\pi}{17} \right) \cdot \: cos \left(  \dfrac{2\pi}{17} \right)  \cdot \:cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{2 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot \:  \left \{2 \:  sin\left(  \dfrac{\pi}{17} \right) cos \left(  \dfrac{\pi}{17} \right)   \right\}\cdot \: cos \left(  \dfrac{2\pi}{17} \right)  \cdot \:cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{2 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot \:  sin\left( \dfrac{2\pi}{17} \right)   \cdot \: cos \left(  \dfrac{2\pi}{17} \right)  \cdot \:cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{4 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot 2\: sin\left( \dfrac{2\pi}{17} \right) \: cos \left(  \dfrac{2\pi}{17} \right)  \cdot \:cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{4 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot \: sin\left( \dfrac{4\pi}{17} \right)   \cdot \:cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{8 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot 2\: sin\left( \dfrac{4\pi}{17} \right)  cos \left(  \dfrac{4\pi}{17} \right)  \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{8 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot \: sin\left( \dfrac{8\pi}{17} \right)   \cdot \: cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{16 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot 2\: sin\left( \dfrac{8\pi}{17} \right)   cos \left(  \dfrac{8\pi}{17} \right)  }

 \sf{ =  \dfrac{1}{16 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot  sin\left( \dfrac{16\pi}{17} \right)   }

 \sf{ =  \dfrac{1}{16 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot  sin\left(  \pi - \dfrac{\pi}{17} \right)   }

 \sf{ =  \dfrac{1}{16 \: sin\left(  \dfrac{\pi}{17} \right) } \cdot  sin\left( \dfrac{\pi}{17} \right)   }

 \sf{ =  \dfrac{1}{16 }  }

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