( sin ⁡ 8+ icos ⁡ 8)^8( sin ⁡ 8- icos ⁡ 8)^8 is equal to

Question

Accepted Answer

Correct answer: C. (c) Given that, ( sin ⁡ 8+ icos ⁡ 8)^8( sin ⁡ 8- icos ⁡ 8)^8= i^8( cos ⁡ 8- isin ⁡ 8)^8 (-i )^8( cos ⁡ 8+ isin ⁡ 8)^8 = ( cos 8 sin ⁡ 8 ) ( cos ⁡ 8 sin ⁡ 8 )= (e^- iπ/8 )^8 (e^ iπ/8 )^8( ∵e^ iθ= cos ⁡ + isin ⁡ ) = e^- iπe^ iπ ( using De- Moivre^ 's Theoraem) = \ cos ⁡ - isin ⁡ \ cos ⁡ + isin ⁡ \= -1-1=1

$\frac{( sin \frac{π}{8} + icos \frac{π}{8} ) ^{8}}{( sin \frac{π}{8} - icos \frac{π}{8} ) ^{8}} is equal to$

Solution

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