IIT JEE , AIEEE Forum - Mathematics , Trignometry - I'm stuck in a simple prooving question, plz. help .

loveshah

Given -:
( (sin ^4

)/a ) + ( ( cos^4

)/b) = 1/(a+b)

Proove - 1)

((sin^8

) /a^3 ) + ( (cos^8

) / b^3) = 1/ (a+b)^3

2)

(sin^4n

) / a^2n-1 + (cos ^4n

) / b^2n-1 = 1 / (a+b)^2n - 1

amar.gupta

Dear,

( (sin ^4

)/a ) + ( ( cos^4

)/b) = 1/(a+b)

let sin ²

= x

then x² /a + (1-x)²/b =1/(a+b)

or bx² + a(1+x²-2x) =ab/(a+b)

or (a+b)x² - 2ax + a²/(a+b) =0

or (a+b)²x² - 2a(a+b) x + a² =0

or [(a+b)x - a]² = 0

or x =a/(a+b) =sin ²

hence cos ²

= b/(a+b)

so now: (sin^4n

) / a^2n-1 + (cos ^4n

) / b^2n-1

= [a/(a+b)]²ⁿ / a^(2n-1)+ [b/(a+b)]²ⁿ / b^(2n-1)or = a/(a+b)²ⁿ + b/(a+b)²ⁿ or = (a+b)/(a+b)²ⁿ or = 1/(a+b)^(2n-1) now put n=2 to get your first part.

gcch29

{(a+b)sin^4(m)/}a + {(a+b)cos^4(m)}/b = 1
= sin^4(m) + cos^4(m) +(b/a)sin^4(m) +(a/b)cos^4(m) = 1
= {(sin^2(m)+co^2(m))^2} - 2sin^2(m)cos^2(m) + [(b/a)^(1/2)sin^2(m) +(a/b)^(1/2)cos^2(m)]
=(b/a)^(1/2)sin^2(m) = (a/b)^(1/2)cos^2(m)
bsin^2(m) = acos^2(m)
sin^2(m)/a = cos^2(m)/b
(sin^2(m)+cos^2(m))/(a+b)=1/(a+b)
therefore
sin^8(m)/a^3 + cos^8(m)/b^3=1/(a+b)^3

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