Trigonometry Proof: Proving Sin^8 + Cos^8 = (a+b)^3

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SUMMARY

The discussion centers on proving the equality \(\frac{sin^8(x)}{a^3} + \frac{cos^8(x)}{b^3} = \frac{1}{(a+b)^3}\) from the initial equation \(\frac{sin^4(x)}{a} + \frac{cos^4(x)}{b} = \frac{1}{a+b}\). Participants attempted various algebraic manipulations, including cubing both sides and cross-multiplying, but encountered difficulties in progressing towards the proof. The consensus suggests that cubing the second equation may be a viable approach to reach the desired conclusion.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sine and cosine functions.
  • Familiarity with algebraic manipulation techniques, including cubing expressions.
  • Knowledge of rational expressions and their properties.
  • Basic skills in mathematical proof techniques, particularly in the context of inequalities and equalities.
NEXT STEPS
  • Research the properties of sine and cosine functions in relation to their powers.
  • Learn about algebraic identities involving cubes and their applications in proofs.
  • Explore advanced trigonometric identities that may simplify the proof process.
  • Study techniques for manipulating rational expressions in mathematical proofs.
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Students studying trigonometry, mathematics enthusiasts, and educators looking to enhance their understanding of algebraic proofs involving trigonometric functions.

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Homework Statement


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

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

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
sin^4(x)+ cos^4(x) + \frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 1

\frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 2sin\frac{x}{2}cos\frac{x}{2}

Cubing this one didnt seem appropriate either. Maybe this is the wrong way I'm going in :-p
Please help..
 
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I think you can cube the 2nd equation :p
 

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