MHB Is This Trigonometric Identity Valid for All Values?

AI Thread Summary
The discussion centers on proving the trigonometric identity involving the equation $\dfrac{\cos^4 a}{x}+\dfrac{\sin^4 a}{y}=\dfrac{1}{x+y}$ for all real values of $a$, $b$, $x$, and $y$. Participants aim to demonstrate that this leads to the conclusion $\dfrac{\cos^8 a}{x^3}+\dfrac{\sin^8 a}{y^3}=\dfrac{1}{(x+y)^3}$. The conversation highlights the mathematical steps and reasoning necessary to validate the identity. The proof requires careful manipulation of the original equation and understanding of trigonometric properties. Overall, the identity is confirmed to hold true under the specified conditions.
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Let $\dfrac{\cos^4 a}{x}+\dfrac{\sin^4 a}{y}=\dfrac{1}{x+y}$ for all real $a,\,b,\,x,\,y$.

Prove that $\dfrac{\cos^8 a}{x^3}+\dfrac{\sin^8 a}{y^3}=\dfrac{1}{(x+y)^3}$
 
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anemone said:
Let $\dfrac{\cos^4 a}{x}+\dfrac{\sin^4 a}{y}=\dfrac{1}{x+y}$ for all real $a,\,b,\,x,\,y$.

Prove that $\dfrac{\cos^8 a}{x^3}+\dfrac{\sin^8 a}{y^3}=\dfrac{1}{(x+y)^3}$

$\dfrac{\cos^4 a }{x} + \dfrac{\sin ^4 a }{y}= \dfrac{1}{x+y}$
hence
$\dfrac{\cos^4 a }{x} + \dfrac{(1-\cos^2 a )^2}{y}= \dfrac{1}{x+y}$
or
$\dfrac{\cos^4 a }{x} + \dfrac{1-2\cos^2 a +cos^4 a}{y}= \dfrac{1}{x+y}$
or
$(x+y)^2\cos^4a-2 x(x +y) \cos^2 a + x(x+y)= xy$
or $(x+y)^2\cos^4a -2 x(x +y) \cos^2 a + x^2= 0$
or $((x+y)\cos^2a -x)^2=0$
hence $\cos^2 a = \dfrac{x}{x+y}\cdots(1)$
from (1)
$\sin ^2 a = 1-\dfrac{x}{x+y}=\dfrac{y}{x+y}\cdots(2)$
using (1) and (2)
$\dfrac{\cos^8 a}{x^3} + \dfrac{\sin ^8 a}{y^3}$
= $\dfrac{(\cos^2 a)^4}{x^3} + \dfrac{(\sin ^2 a)^4}{y^3}$
= $\dfrac{(\frac{x}{x+y})^4}{x^3} + \dfrac{(\frac{y}{x+y})^4}{y^3}$
= $\dfrac{x}{(x+y)^4} + \dfrac{y}{(x+y)^4}$
= $\dfrac{x+y}{(x+y)^4}$
= $\dfrac{1}{(x+y)^3}$
 
Good job, kaliprasad!:cool:
 
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