Cos8-7cos10-7cos12-7 is equal to

The correct option is B 1/8 cos 8π #10; #160; 7 cos 10π #160; 7 cos 12π #160; 7 #10; =cos 8π #160; 7 cos( 2π 4π #160; 7 #10; ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

cos8π/7cos10π...

Question

$cos \frac{8 π}{7} cos \frac{10 π}{7} cos \frac{12 π}{7}$ is equal to

\frac{1}{2}

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\frac{- 1}{2}

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\frac{- 1}{8}

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\frac{1}{4}

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Solution

The correct option is B $\frac{- 1}{8}$
$cos \frac{8 π}{7} cos \frac{10 π}{7} cos \frac{12 π}{7} = cos \frac{8 π}{7} cos (2 π - \frac{4 π}{7}) cos (2 π - \frac{2 π}{7}) = cos \frac{8 π}{7} cos \frac{4 π}{7} cos \frac{2 π}{7} = \frac{2 sin \frac{2 π}{7} cos \frac{2 π}{7} cos \frac{4 π}{7} cos \frac{8 π}{7}}{2 sin \frac{2 π}{7}} = \frac{2 sin \frac{4 π}{7} cos \frac{4 π}{7} cos \frac{8 π}{7}}{4 sin \frac{2 π}{7}} = \frac{2 sin \frac{8 π}{7} cos \frac{8 π}{7}}{8 sin \frac{2 π}{7}} = \frac{sin \frac{16 π}{7}}{8 sin \frac{2 π}{7}} = \frac{sin (2 π + \frac{2 π}{7})}{8 sin \frac{2 π}{7}} = - \frac{sin \frac{2 π}{7}}{8 sin \frac{2 π}{7}} = - \frac{1}{8}$

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Similar questions

Q. Prove that

cos \frac{2 π}{7} . cos \frac{4 π}{7} . cos \frac{8 π}{7} = \frac{1}{8}

$\frac{1}{(\sqrt{9} - \sqrt{8})} - \frac{1}{(\sqrt{8} - \sqrt{7})} + \frac{1}{(\sqrt{7} - \sqrt{6})} - \frac{1}{(\sqrt{6} - \sqrt{5})} + \frac{1}{(\sqrt{5} - \sqrt{4})}$ is equal to:

2^{1 / 4} \cdot 4^{1 / 8} . 8^{1 / 16} . 16^{1 / 32} . . . .

is equal to

2^{1 / 4} {.4}^{1 / 8} {.8}^{1 / 16} {.16}^{1 / 32}

....... is equal to

$\frac{1}{4} \times (1 + \frac{1}{2}) \times (1 + \frac{3}{4}) \times (1 + \frac{5}{4}) \times (1 + \frac{7}{4}$ ) is equal to:

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cos8π7cos10π7cos12π7 is equal to

The correct option is B −18cos8π7cos10π7cos12π7=cos8π7cos(2π−4π7)cos(2π−2π7)=cos8π7cos4π7cos2π7=2sin2π7cos2π7cos4π7cos8π72sin2π7=2sin4π7cos4π7cos8π74sin2π7=2sin8π7cos8π78sin2π7=sin16π78sin2π7=sin(2π+2π7)8sin2π7=−sin2π78sin2π7=−18

$cos \frac{8 π}{7} cos \frac{10 π}{7} cos \frac{12 π}{7}$ is equal to