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Math Help - Trigonometry

  1. #1
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    Trigonometry

    Show that \sin\left[\frac{\pi}{14}\right]\cdot\sin\left[\frac{3\pi}{14}\right]\sin\left[\frac{5\pi}{14}\right]=\frac 18.
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  2. #2
    Junior Member AlvinCY's Avatar
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    Would this happen to be in the "complex numbers" section of the textbook/exercise? If so, did it have a first part? Also, is that the question? Or does it say "hence show that..."

    Because I have seen these questions in our 4U textbooks (highest level maths in high school), and recognise them, I might be able to give you a hand if there is a first part!
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  3. #3
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    I'd appreciate your help. There are no other parts. This is the only question but I don't know from where did they get this.. It's between a collection of exercises xD and no complex numbers nearby.
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  4. #4
    MHF Contributor red_dog's Avatar
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    8\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\  pi}{14}=4\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)\sin\frac{5\pi}{14}=

    2\left(2\sin\frac{5\pi}{14}\cos\frac{\pi}{7}-2\sin\frac{5\pi}{14}\cos\frac{2\pi}{7}\right)=2\le  ft(\sin\frac{\pi}{2}+\sin\frac{3\pi}{14}-\sin\frac{9\pi}{14}-\sin\frac{\pi}{14}\right)=

    =2\left(1+\sin\frac{3\pi}{14}-\sin\left(\pi-\frac{9\pi}{14}\right)-\sin\frac{\pi}{14}\right)=2\left(1+\sin\frac{3\pi}  {14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}\right)

    Now I will show that \sin\frac{3\pi}{14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}=-\frac{1}{2} (*)

    This is equivalent to \sin\frac{\pi}{14}-\sin\frac{3\pi}{14}+\sin\frac{5\pi}{14}=\frac{1}{2  }


    Multiply both members by 2\sin\frac{\pi}{7} then we transform products to sums.

    \cos\frac{\pi}{14}-\cos\frac{3\pi}{14}-\cos\frac{\pi}{14}+\cos\frac{5\pi}{14}+\cos\frac{3  \pi}{14}-\cos\frac{\pi}{2}=\sin\frac{\pi}{7}

    \cos\frac{5\pi}{14}=\sin\frac{\pi}{7}

    \cos\frac{5\pi}{14}=\cos\left(\frac{\pi}{2}-\frac{\pi}{7}\right)

    \cos\frac{5\pi}{14}=\cos\frac{5\pi}{14}


    So, the identity (*) is true.
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