if (sin^4x)/a + (cos^4x)/b = (1/(a +b)) Prove that (sin^3x)/a^3 + (cos^3x)/b^3 = (1/(a +b)^3)
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if (sin^4x)/a + (cos^4x)/b = (1/(a +b))
Prove that (sin^8x)/a + (cos^8x)/b = (1/(a +b)^3)
7 years ago
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should be corrected as shown
Q.
if (sin^4x)/a + (cos^4x)/b = (1/(a +b))
Prove that (sin^8x)/a^3 + (cos^8x)/b^3 = (1/(a +b)^3)
7 years ago
cos^4x=(cos^2x)^2=(1-sin^2x)^2=1+ sin^4x-2sin^2x
=>sin^4x / a +cos^4x / b=1/a+b
=>sin^4x/a + (1+sin^4x-2sin^2x)/b = 1/(a+b)
[b*sin^4x + a(sin^4x-2sin^2x+1)] /ab = 1/(a+b)
=>[(a+b)sin^4x-2a sin^2x+a]/ab = 1/a+b
=>(a+b)^2 sin^4x - 2a(a+b)sin^2x + a(a+b) =ab
=>(a+b)^2 sin^4x - 2a(a+b)sin^2x + a^2
=> [(a+b)sin^2x-a]^2 = 0
=>(a+b)sin^2x - a = 0
sin^2x=a/(a+b).........(1)
(take fourth power of both side)
=>sin^8x=a^4/(a+b)^4
(divide by a^3 both side)
=>sin^8x/a^3=a/(a+b)^4.
........(2)
=>cos^2x=1 - sin^2x=1-a/(a+b)=b/(a+b). (from eq 1 substituting value of sin^2x)
=>cos^2x=b/(a+b)...........(3)
(take fourth power of both side)
=>cos^8x=b^4/(a+b)^4
(divide by b^3 both side)
=>cos^8x/b^3 =b/(a+b)^4......(4)
(adding eq 2@4)
=>sin^8x/a^3 + cos^8x/b^3=a/(a+b)^4 + b/(a+b)^4 =(a+b)/(a+b)^4=1/(a+b)^3 ........proved
3 years ago sin^4x /a + cos^4x /b =1/a+ba+b (bsin^4x +acos^4x )=ababsin^4x +b^2sin^4x +abcos^4x +a^2cos^4x =abab (sin^4x +cos^4x )+a^2cos^4x +b^2sin^4x =abab (1-2sin^2x.cos^2x )+a^2cos^4x +b^2sin^4x =aba^2cos^4x +b^2sin^4x -2absin^2x.cos^2x=0(acos^2x+bsin^2x)^2=0acos ^2x=bsin^2xa (1-sin^2x)=bsin^2xa-asin^2x=sin^2xa=(a+b)sin^2xsin^2x=a/a+b (1)similarly......cos^2x=b/a+b (2)take 4th root both side in both eq..then we have.....Sin^8x=a^4/(a+b)^4 (3)cos^8x=b^4/(a+b)^4 (4)now divide eq (3)by a^3 and eq (4) byb^3and then add the eq..after doing that we have .....sin^8x/a^3 +cos^8x/b^3 =a/(a+b)^4 +b/(a+b)^4sin^8x/a^3 +cos^8x/b^3 =1/(a+b)^3hence proved