Trigonometry proof

  • Thread starter The legend
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  • #1
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Homework Statement


Given the equality
[tex]\frac{sin^4(x)}{a} + \frac{cos^4}{b} = \frac{1}{a+b}[/tex]

Prove that :
[tex]\frac{sin^8(x)}{a^3} + \frac{cos^8}{b^3} = \frac{1}{(a+b)^3}[/tex]

The Attempt at a Solution


I cubed on both the sides of the 1st equation and solved a bit, reaching no where. Then I tried by cross multiplying a+b, getting
[tex]sin^4(x)+ cos^4(x) + \frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 1[/tex]

[tex]\frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 2sin\frac{x}{2}cos\frac{x}{2}[/tex]

Cubing this one didnt seem appropriate either. Maybe this is the wrong way i'm going in :tongue:
Please help..
 

Answers and Replies

  • #2
559
8
I think you can cube the 2nd equation :p
 

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