Cos2 -7-cos4 -7-cos6 -7-

The correct option is D Roots of z^7 = 1 are e^2kπ i/7,where k = 0,1,2,.....,6. Sum of roots = 0 ⇒ nbsp;Real part of sum of roots = 0 ⇒ 1 + cos2π/7 + cos4π ...

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Byju's Answer

Standard XI

Mathematics

Equations of the form acos?+bsin? = c

cos2 π/7+cos4...

Question

$c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7}$ =

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Solution

The correct option is D

Roots of $z^{7} = 1$ are $e^{\frac{2 k π i}{7}}$ ,where k = 0,1,2,.....,6.

Sum of roots = 0 $\Rightarrow$ Real part of sum of roots = 0

$\Rightarrow 1 + c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7} + c o s \frac{8 π}{7} + c o s \frac{10 π}{7} + c o s \frac{12 π}{7} = 0$

$\Rightarrow c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7} = - \frac{1}{2}$

Suggest Corrections

cos2π7+cos4π7+cos6π7 =

The correct option is D Roots of z7=1 are e2kπi7,where k = 0,1,2,.....,6. Sum of roots = 0 ⇒ Real part of sum of roots = 0 ⇒1+cos2π7+cos4π7+cos6π7+cos8π7+cos10π7+cos12π7=0 ⇒cos2π7+cos4π7+cos6π7=−12

$c o s \frac{2 π}{7} + c o s \frac{4 π}{7} + c o s \frac{6 π}{7}$ =