Trigonometry> proof required...

(a +b)^3)

sshanbhag_2000@yahoo.com

Subhash Shanbhag , 14 Years ago

Grade 12

6 Answers

Subhash Shanbhag

the question has an error

should be corrected as shown

if (sin^4x)/a + (cos^4x)/b = (1/(a +b))

Prove that (sin^8x)/a^3 + (cos^8x)/b^3 = (1/(a +b)^3)

Last Activity: 14 Years ago

Shamik Banerjee

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

Last Activity: 10 Years ago

Sahil Saini

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

Last Activity: 8 Years ago

Yash verma

Sin^4x/a +cos^4x/b =1/(a+b)....eq(1)(1-cos^2x)b + cos^4xa =ab/(a+b)b+bcos^4x-2bcos^2x+cos^4ax-ab/(a+b)=0Cos^4x(a+b)-2bcos^2x +b-ab/(a+b)=0Cos^4x(a+b)^2-2b(a+b)cos^2x+b^2=0Cos^2x= (2b(a+b)+-√((4b^2(a+b)^2-4b^2(a+b)^2))/2(a+b)=b/(a+b)------by quadratic equationSo, cos^4x=b^2/(a+b)^2And sin^4xb=ab/(a+b) - cos^4xa by eq(1)so sin^4x=a^2/(a+b)^2Putting sin^8x/b^3+cos^8x/a^3, the values of sin and cos we get,(a+b)/(a+b)^4=1/(a+b)^3

Last Activity: 8 Years ago

Anonymous

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Last Activity: 8 Years ago

Yash Chourasiya

Dear Student

sin⁴x/a + cos⁴ x /b = 1/(a + b)

=> sin⁴ x/a + (cos² x)²/b = 1/(a + b)

=>( sin² x)²/a + (1 - sin²x)²/b = 1/(a + b)

Let, sin² x = k

k²/a + (1 - k)²/b = 1/(a + b)

=> [bk² + a(1 - k)²]/ab = 1/(a+ b)

=> [bk² +a(1 - 2k + k²)]/ab = 1/(a + b)

=> [bk² + a - 2ak + ak²]/ab = 1/(a + b)

=> [k²(a + b) - 2ak + a]/ab = 1/(a + b)

=> k²(a + b)² - 2a(a + b)k + a(a + b) - ab = 0

=> k²(a+ b)² - 2a(a + b)k + a² + ab - ab = 0

=> k²(a + b)² - 2a(a + b)k + a² = 0

=> {k(a + b)}² - 2.k(a + b).a + (a)² = 0

=> [k(a+ b) - a]² = 0

=> k(a + b) - a = 0

=> k(a + b) = a

=> k = a/(a + b)

Hence,

sin² x = a/(a+ b)

cos² x = 1 - a/(a + b) =[a + b - a] /(a + b) = b/(a+ b)

Now,
sin⁸ x = (sin² x)⁴ = [a/(a + b)]⁴ = a⁴/(a + b)⁴

=> sin⁸ x/a³ = a⁴/(a + b)⁴.a³ = a/(a + b)⁴

Similarly, cos⁸ x = (cos² x)⁴ = [b/(a + b)]⁴
= b⁴/(a + b)⁴

=> cos⁸ x/b³ = b⁴/(a + b)⁴.b³ = b/(a + b)⁴

Now, sin⁸ x/a³+ cos⁸x /b³

= a/(a+ b)⁴ + b/(a + b)⁴

= (a + b)/(a+ b)⁴

= 1/(a + b)³
Hence Proved.

I hope this anwer will help you.
Thanks & Regards
Yash Chourasiya

Last Activity: 5 Years ago