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Sum of Trigonometric Ratios in Terms of Their Product

The value of ...

Question

The value of ${(sin \frac{π}{8} + i cos \frac{π}{8})}^{8}$

A

-1

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B

1

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C

0

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D

None of these

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Solution

The correct option is A -1
$z = {(sin \frac{π}{8} + i cos \frac{π}{8})}^{8} = [i (cos \frac{π}{8} - i sin \frac{π}{8})]]^{8}$
...{ $∵ i^{8} = 1$ }
$\Rightarrow z = cos π - i sin π = - 1$ ...{De Moivre's Theorem}
Hence, option 'A' is correct.

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Sum of Trigonometric Ratios in Terms of Their Product

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