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Trignometry

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9 Mar 2008 10:22:14 IST

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Trig Equation

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$ext {This is a 10th Class problem} ext {If} rac{sin^4 heta} {a} + rac{cos^4 heta} {b} = rac{1}{a+b} ext{then prove that} rac{sin^8 heta} {a^3} + rac{cos^8 heta} {b^3} = rac{1}{(a+b)^3} ext{I would like to see the shortest approach to this one}$

Note: a,b are positive numbers

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Comments (4)

kaushik krishna

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9 Mar 2008 10:33:54 IST

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mr. bhatt this was the shortest method i cud think of....

subst..it seems to work out evry time....

pl. wait i wud like to give a better proof

that smiley at the end gives me a gut feeling the method wud b simple

anchit saini

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9 Mar 2008 10:50:42 IST

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using lemma which states that
x^2/a + y^2/b>=(x+y)^2/(a+b) [equality when x/a=y/b]

we get from the first eqn

(sin^2 theta)^2/a + (cos^2 theta)^2/b >=
(sin^2 theta + cos^2 theta)^2/(a+b)=1/(a+b)

since it is the case of equality--
sin^2 theta / a = cos^2 theta / b

now we can put any value of theta a, b to fit the above condition
with the best being
theta=45 and a=b=1

from this we easily get that--
sin^8 theta/a^3 + cos^8 theta / b^3 =1/16+1/16=1/8 = 1/(a+b)^3

this method seems much longer in written form than it is in while doing mentally!!
i know sir this is not the method u are looking for but it still seems ok!!!

Hari Shankar

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9 Mar 2008 11:16:14 IST

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Anchit, that is the soln I had in mind, except the last part:

$\frac {a^2} {x} + \frac {b^2} {y} \geq \frac {(a+b)^2} {x+y} \\ \\ \text {the equality occurs under the condition} \ \frac {a} {x} = \frac {b} {y} \\ \\ \text {Hence} \ \frac{\sin^4\theta} {a} + \frac{\cos^4\theta} {b} \geq \frac {(\sin^2\theta + \cos^2\theta)^2} {a+b} = \frac {1} {a+b} \\ \\ \text {Since equality holds, we must have} \\ \\ \frac {\sin^2\theta} {a} = \frac {\cos^2\theta} {b} \\ \\ \text {or} \ \tan^2\theta = \frac {a} {b} \\ \\ \text {This gives} \ \sin^2\theta = \frac {a} {a+b} \ \text {and} \ \cos^2\theta = \frac {b}{a+b} \\ \\ \text {Hence,}\ \frac{\sin^8\theta} {a^3} + \frac{\cos^8\theta} {b^3} = \sin^2\theta \ (\frac {\sin^2\theta} {a})^3 + \cos^2\theta \ (\frac {\cos^2\theta} {b})^3 \\ \\ = \ \sin^2\theta \ \frac{1} {(a+b)^3} + \cos^2\theta \ \frac{1} {(a+b)^3} \\ \\ = \ \frac{1}{(a+b)^3} \\ \\$

anchit saini

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9 Mar 2008 11:20:34 IST

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sir i think that
that shortest approach u wrote after ur qn made me take a real shortcut!!!!

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