Sin⁴∅/x + Cos⁴∅/y=1/x+y then prove sin¹²∅/x to the power 5 + cos¹²∅/y to the power 5=1/(x+y) to the power 5?

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Answer 1 · 2018-04-01T06:52:54

Kindly refer to a Proof in the Explanation.

Let, $cos^2phi=a," so that, "sin^2phi=1-a$ .

Given that, $cos^4phi/y+sin^4phi/x=1/(x+y)$ ,

$:. a^2/y+(1-a)^2/x=1/(x+y)$ .

$:. {a^2x+(1-a)^2y}/(xy)=1/(x+y)$ .

$:. {a^2x+(1-2a+a^2)y}/(xy)=1/(x+y)$ .

$:. {a^2(x+y)-2ay+y}/(xy)=1/(x+y), i.e.,$

$a^2(x+y)^2-2ay(x+y)+y(x+y)=xy, or,$

$a^2(x+y)^2-2ay(x+y)+y^2=0$ .

$:. {a(x+y)-y}^2=0$ .

$rArr a=y/(x+y), i.e., cos^2phi=a=y/(x+y)$ .

$:. sin^2phi=1-a=1-y/(x+y)=x/(x+y)$ .

In other words, $cos^2phi/y=1/(x+y), &, sin^2phi/x=1/(x+y)$ .

Accordingly, $(cos^2phi/y)^6=1/(x+y)^6, i.e.,$

$cos^12phi/y^6=1/(x+y)^6 :. cos^12phi/y^5=y/(x+y)^6$ .

Similarly, $sin^12phi/x^5=x/(x+y)^6$ .

Consequently, $cos^12phi/y^5+sin^12phi/x^5$ ,

$=y/(x+y)^6+x/(x+y)^6=1/(x+y)^6(y+x)$ .

$rArr cos^12phi/y^5+sin^12phi/x^5=1/(x+y)^5$ , as desired!

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