Cos π/7*Cos 2π/7*Cos 10π/7 – Sin π/14*Sin 3π/14*Sin 5π/14 =
Cos π/7*Cos 2π/7*Cos 10π/7 – Sin π/14*Sin 3π/14*Sin 5π/14 =
Shubham Sahoo , 7 Years ago
Grade 12th pass
IIT JEE Entrance Exam> Cos π/7*Cos 2π/7*Cos 10π/7 – Sin π/14*Sin...
1 Answers
Arun
Last Activity: 5 Years ago
E =sin(π/14) sin(3π/14) sin(5π/14)
Put : x = π/14.
Then : 14x = π.
∴ E = sin x. sin 3x, sin 5x.
∴ E = sin (π/14). sin (3π/14). sin (5π/14) ......... (1)
Now,
sin π/14 = sin (π/2 - 6π/14) = cos 6π/14 = cos (π - 8π/14) = - cos 8π/14 = - cos 8x
Similarly, sin 3π/14 = cos 4π/14 = cos 4x and
sin 5π/14 = cos 2π/14 = cos 2x.
∴ from (1),
E = - cos 8x. cos 4x. cos 2x
Multiply and divide by 2 sin2x,
...= -1/ (2 sin2x ). [( 2 sin 2x. cos 2x ). cos 4x. cos 8x]
...= (-1/ 4 sin 2x ). ( 2 sin 4x. cos 4x ). cos 8x
...= (-1/ 8 sin 2x ). ( 2 sin 8x. cos 8x )
...= (-1/ 8 sin 2x ). sin 16x
...= (-1/8). ( 1/ (sin π/7)). sin 8π/7
...= (-1/8). ( 1/ (sin π/7)). sin ( π + π/7 )
...= - (1/8). ( 1/ sin π/7 ). ( - sin π/7 )
...= 1/8.
∴ E = 1/8 is the answer
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