If cos 2-3-cos -cos 4-3-cos 5-3-cos 2- -cos7-3- -cos40-3-cos41-3-k- then the value of -6k- is

Given, #160;k=cos2π/3 cos (2π/3+π/3 )+cos (2π/3+2.π/3 ) cos (2π/3+3π/3 )+.... cos (2π/3+39.π/3 ) #160;=cos2π/3+cos (2π/3+π+π/3 )+cos (2π/3+2 (π+π/3 ) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

If cos 2π/3...

Question

If $c o s \frac{2 π}{3} - c o s π + c o s \frac{4 π}{3} - c o s \frac{5 π}{3} + c o s 2 π - c o s \frac{7 π}{3} + . . . . + c o s \frac{40 π}{3} - c o s \frac{41 π}{3} = k$ , then the value of |6k| is

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Solution

Given, $k = c o s \frac{2 π}{3} - c o s (\frac{2 π}{3} + \frac{π}{3}) + c o s (\frac{2 π}{3} + 2. \frac{π}{3}) - c o s (\frac{2 π}{3} + \frac{3 π}{3}) + . . . . - c o s (\frac{2 π}{3} + 39. \frac{π}{3})$
$= c o s \frac{2 π}{3} + c o s (\frac{2 π}{3} + π + \frac{π}{3}) + c o s (\frac{2 π}{3} + 2 (π + \frac{π}{3})) + . . . . + c o s (\frac{2 π}{3} + 39 (π + \frac{π}{3}))$
$∴ α = \frac{2 π}{3}, β = \frac{4 π}{3}$ and $n = 40$
$k = \frac{s i n (\frac{40}{2} . \frac{4 π}{3}) [c o s (\frac{2 π}{3} + \frac{39}{2} . \frac{4 π}{3})]}{s i n (\frac{4 π}{6})}$
$= \frac{s i n (\frac{80 π}{3}) c o s (\frac{80 π}{3})}{s i n (\frac{2 π}{3})}$ $= \frac{s i n (\frac{160 π}{3})}{2 s i n (\frac{2 π}{3})} = \frac{s i n (53 π + \frac{π}{3})}{2 s i n (\frac{2 π}{3})} = - \frac{1}{2}$
$\Rightarrow | 6 k | = ∣ ∣ ∣ 6. (- \frac{1}{2}) ∣ ∣ ∣ = 3$

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If cos2π3−cosπ+cos4π3−cos5π3+cos2π−cos7π3+....+cos40π3−cos41π3=k, then the value of |6k| is

If $c o s \frac{2 π}{3} - c o s π + c o s \frac{4 π}{3} - c o s \frac{5 π}{3} + c o s 2 π - c o s \frac{7 π}{3} + . . . . + c o s \frac{40 π}{3} - c o s \frac{41 π}{3} = k$ , then the value of |6k| is