This question is related to trigonometry.pls solve as fast as possible and send solution .
This question is related to trigonometry.pls solve as fast as possible and send solution .
Ritheesh , 7 Years ago
Grade 11
Trigonometry> This question is related to trigonometry....
1 AnswersAditya Gupta
ok so the question you have asked is a wonderful ques no doubt.
let beta=b. now, i am sure you are familiar with the identity cotx – tanx= 2cot2x
so the general term of the summation becomes 2^r*tan2^rb = 2^r*(cot2^rb – 2cot2^(r+1)b)
= 2^rcot2^rb – 2^(r+1)cot2^(r+1)b which is TELESCOPIC
so ∑r=0 to n – 1 2^rcot2^rb – 2^(r+1)cot2^(r+1)b = cotb – 2^ncot2^nb
or tanb+2tan2b+...+2^n-1tan2^n-1b+ 2^ncot2^nb= cotb
so now all we gotta do is to find the value of b.
given that cos(a)+cos(a+b)+cos(a+b+y)=0
and sin(a)+sin(a+b)+sin(a+b+y)=0
multiply second eqn by iota and add the first eqn to it to get
e^ia+e^i(a+b)+e^i(a+b+y)=0
divide by e^ia to get
1+e^ib+e^i(b+y)=0
similarly, multiply second eqn by minus iota and add the first eqn to it to get
e^(-ia)+e^–i(a+b)+e^–i(a+b+y)=0
multiply by e^ia to get
1+e^–ib+e^–i(b+y)=0
1+e^–ib+e^–i(b+y)=0
let e^ib = x and e^i(b+y) = z
then x+z+1=0 and 1+1/x+1/z=0
eliminating z from these eqns, we have
x^2+x+1=0
so x= omega, omega square= e^i(2pi/3) or e^i(4pi/3)
but since b belongs to (0, pi)
so x= e^i(2pi/3)
so e^ib=e^i(2pi/3)
or b=2pi/3
so that sum= cotb= cot2pi/3
= – 1/√3
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