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Compare the numbers 3tan 2 and 2tan 3
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We can compare the two terms by using Lagrange's theorem which gives that a function is increasing if the first derivative is positive.
Take the function tan x / x, here x lies in the interval (0, 3*pi/180)
Differentiating the function using the quotient rule we get:
[tan x/x]' = [(x/(cos x)^2 - sin x/cos x]/x^2
=> (2x - 2* sin x cos x)/2*x^2*(cos x)^2
=> (2x - sin 2x)/2*x^2*(cos x)^2
As (2x - 2* sin x cos x)/2*x^2*(cos x)^2 > 0, the function f(x) is increasing.
3*pi/180 > 2*pi/180
=> f(3*pi/180) > f(2*pi/180)
pi radians = 180 degrees
=> (tan 3) / 3 > (tan 2) / 2
=> 2*tan 3 > 3*tan 2
This gives 2*tan 3 > 3*tan 2
Posted by justaguide on April 28, 2011 at 12:40 AM (Answer #1)
We'll choose a function whose domain of definition is (0 ; 3pi/180).
The function is f(x) = (tan x)/x
We'll do the first derivative test to verify the if the function is monotonic. We'll apply the quotient rule:
f'(x) = [(tan x)'*x - (tan x)*(x)']/x^2
f'(x) = [x/(cos x)^2 - sin x/cos x]/x^2
f'(x) = (x - sin x*cos x)/x^2*(cos x)^2
We notice that the numerator is positive for any value from the domain of definition and the denominator is always positive.
Then f'(x)>0 => the function is monotonically increasing.
If so, for x=2pi/180 < x=3pi/180 => f(2pi/180) < f(3pi/180)
But f(2pi/180) = tan (2pi/180)/(2pi/180)
f(3pi/180) = tan (3pi/180)/(3pi/180)
tan (2pi/180)/(2pi/180) < tan (3pi/180)/(3pi/180)
We'll simplify by pi/180:
(tan 2)/2 < (tan 3)/3
We'll cross multiply and we'll get:
3*tan 2 < 2*tan 3
Comparing the given values, we have found the inequality: 3*tan 2 < 2*tan 3.
Posted by giorgiana1976 on April 28, 2011 at 12:41 AM (Answer #2)
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