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Range of Trigonometric Ratios from 0 to 90 Degrees

Prove that: ...

Question

Prove that:
$sin \frac{π}{15} + sin \frac{4 π}{15} - sin \frac{11 π}{15} - sin \frac{14}{15} = 0$

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Solution

$L H S = s i n \frac{π}{15} + s i n \frac{4 π}{15} - s i n \frac{11 π}{15} - s i n \frac{14 π}{15}$

$L H S = s i n \frac{π}{15} + s i n \frac{4 π}{15} - s i n (π - \frac{11 π}{15}) - s i n (π - \frac{14 π}{15})$ [Since $s i n (π - x) = s i n x$ ]

$L H S = s i n \frac{π}{15} + s i n \frac{4 π}{15} - s i n \frac{4 π}{15} - s i n \frac{π}{15}$

$L H S = 0.$

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sin \frac{π}{15} + sin \frac{4 π}{15} - sin \frac{11 π}{15} - sin \frac{14 π}{15} = 0

sin \frac{π}{15} + sin \frac{4 π}{15} - sin \frac{11 π}{15} - sin \frac{14 π}{15} = m

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Q. Prove that :
4cos2A
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\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15} = \frac{1}{16}

\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{3 π}{15} \cos \frac{4 π}{15} \cos \frac{5 π}{15} \cos \frac{6 π}{15} \cos \frac{7 π}{15} = \frac{1}{128}

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