24sir isn't a term missing ?????becoz.......
1,3,5,7,9,_,13
Evaluate: \sin \left(\frac{\pi}{14} \right)\sin \left(\frac{3\pi}{14} \right)\sin \left(\frac{5\pi}{14} \right)\sin \left(\frac{7\pi}{14} \right)\sin \left(\frac{9\pi}{14} \right)\sin \left(\frac{13\pi}{14} \right).sin(11pi/14)
Today I am on an IIT JEE past papers frenzy!
This will last for 1 week atleast..
please bear with me ;)
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14 Answers
62but then solve these too..
I am only trying to show how simple JEE questions generally are! :)
11the ques is
cos6pi/14 x cos4pi/14 x cos2pi/14 x sin5pi/14 x sin pi/14
multiply upper niiche by 2 to make the required form[1]
nw apply formula of sina + sinb =2sina+b/2 cos a+b/2
106\sin (\pi /14) = sin(\pi /2 -3\pi/7) = cos(3\pi /7)
\sin (3\pi /14) = sin(\pi /2 -2\pi/7) = cos(2\pi /7)
\sin (5\pi /14) = sin(\pi /2 -\pi/7) = cos(\pi /7)
\sin (7\pi /14) = sin(\pi /2) = 1
\sin (9\pi /14) = sin(\pi /2 +\pi/7) = cos(\pi /7)
\sin (11\pi /14) = sin(\pi /2 +2\pi/7) = cos(2\pi /7)
\sin (13\pi /14) = sin(\pi /2 +3\pi/7) = cos(3\pi /7)
cos(3\pi /7) = -cos(\pi -4\pi /7) = -cos(4\pi /7)
so the final equation becomes
[cos(3\pi /7)*cos(2\pi /7)*cos(\pi /7)]^2 = [cos(\pi /7)*cos(2\pi /7)*cos(4\pi /7)]^2
106= [sin(8Ï€/7)/8*sin(Ï€/7)]2
= 1/64
[calculation errors might be there]
106i know .. i accidentally pressed add post when i still had the last step to write...