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Grade: 12th pass
        
(13)Let k=1°, then prove that sigma n=0 to 88 1/cosnk×cos(n+1)k = cosk/sin^2 k
(14)let x belongs to R and sigma k=2 to infinity sin(2^k x) = a. Find the value of sigma k=0 to infinity [cot^3 (2^k x) - cot (2^k x)]sin^4(2^k x) in term of 'a'
(15) given that 3 sinx +4 coax =5 where x belongs to (0,π/2). Find the value of 2 sinx + coax + 4tanx
10 months ago

Answers : (2)

Deepak Kumar Shringi
askIITians Faculty
4370 Points
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10 months ago
Aditya Gupta
1124 Points
							
kindly dont ask so many questions in one go, you can instead ask 3 separate questions.
  1. let S= ∑1/cosnk×cos(n+1)k, so S*sink= ∑(sin((n+1)k-nk)/cosnk×cos(n+1)k=∑tan(n+1)k-tan(nk) [using sin(A-B)=sinAcosB-sinBcosA]. the above sum now telescopes, giving us  S*sink= tan89-tan0= sin89/cos89=cos1/sin1. therefore, S= cosk/sin^2k HP
  2.  let us assume that 2^k*x=y. so that ∑ [cot^3 (2^k x) - cot (2^k x)]sin^4(2^k x) =∑ [cos^3(y)/sin^3y – cosy/siny]sin^4y= ∑cos^3ysiny-cosysin^3y=∑cosysiny(cos^2y – sin^2y)= (∑sin4y)/4=(∑sin(2^(k+2)x)/4= (a – sinx – sin2x)/4
  3. (3/5)sinx+(4/5)cosx= 1 let cosa=3/5 so that sina=4/5. implies sinxcosa+sinacosx=1 or sin(x+a)=sin 90deg or x+a= 90deg, so sinx=sin(90 – a)=cosa=3/5 and cosx=4/5 and tanx=3/4 so that 2 sinx + coax + 4tanx= 2+3= 5
10 months ago
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