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Compare the numbers 3tan 2 and 2tan 3

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sodelete | Student, College Freshman | eNoter

Posted April 28, 2011 at 12:11 AM via web

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Compare the numbers 3tan 2 and 2tan 3

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 28, 2011 at 12:40 AM (Answer #1)

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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

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giorgiana1976 | College Teacher | Valedictorian

Posted April 28, 2011 at 12:41 AM (Answer #2)

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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.

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