The value of -k-16sin2 - k-7-icos2 - k-7 is-

The correct option is B 1 Given expression ∏_k=1^6 ( i) ( cos 2π k/7 + i sin2π k/7 ) ( i)^6 ( cos 2π k/7 + i sin2π k/7 ) × ( cos 4π/7 + i sin4π/7 ).....6term ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Equations

The value of ...

Question

The value of $\prod_{k = 1}^{6} (s i n \frac{2 π k}{7} - i c o s \frac{2 π k}{7})$ is:

-1

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-i

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Solution

The correct option is A
-1

Given expression $\prod_{k = 1}^{6} (- i) (c o s \frac{2 π k}{7} + i s i n \frac{2 π k}{7})$

$(- i)^{6} (c o s \frac{2 π k}{7} + i s i n \frac{2 π k}{7})$ $\times$ $(c o s \frac{4 π}{7} + i s i n \frac{4 π}{7}) . . . . .6 t e r m s$

= - $[c o s (\frac{2 π}{7} + \frac{4 π}{7} + . . . . . . .6 t e r m s) + i s i n (\frac{2 π}{7} + \frac{4 π}{7} + . . . . . . .6 t e r m s)]$

= $- [c o s (\frac{42 π}{7}) + i s i n (\frac{42 π}{7})]$

= - $(c o s 6 π + i s i n 6 π) = - 1$

Suggest Corrections

The value of ∏6k=1(sin2πk7−icos2πk7) is:

The correct option is A -1 Given expression ∏6k=1(−i)(cos2πk7+isin2πk7) (−i)6(cos2πk7+isin2πk7) × (cos4π7+isin4π7).....6terms = -[cos(2π7+4π7+.......6terms)+isin(2π7+4π7+.......6terms)] = −[cos(42π7)+isin(42π7)] = -(cos6π+isin6π)=−1

The value of $\prod_{k = 1}^{6} (s i n \frac{2 π k}{7} - i c o s \frac{2 π k}{7})$ is: