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Compare 3*tg 2 to 2*tg 3!

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radreala | Student, College Freshman | (Level 3) eNoter

Posted April 28, 2009 at 11:24 PM via web

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Compare 3*tg 2 to 2*tg 3!

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted May 25, 2009 at 5:58 PM (Answer #1)

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We'll use a consequence of Lagrange's theorem which says that a function is increasing if it's first derivative is positive.

Let's consider a function f(x)=tg x/x, where x is in the set       (0, 3pi/180).

To analyze if we have an increasing function, we have to calculate the first derivative. We've noticed that the function is a fraction and we'll calculate it's derivative following this rule:

(f/g)'= (f'*g-f*g')/g^2

(tg x/x)'= [(tg x)'*x-(tgx)*x']/x^2

(tg x/x)' = [(x/cos^2 x) - sin x/cosx]/x^2

(tg x/x)' = (2x-2sinx cosx)/2*x^2*cos^2x

It's obvious that (2x - sin 2x)/2*x^2*cos^2x>0, so f(x) is increasing, so that, for 2pi/180<3pi/180,

f(2pi/180)<f(3pi/180),

tg(2pi/180)/2pi/180<tg(3pi/180)/3pi/180

pi(radians)=180(degrees) represents the same measure of an angle, so we'll reduce pi with 180

tg(2)/2<tg(3)/3

We'll cross multiplying and the result will be

3tg(2)<2tg(3)

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