Trig Function Inequality Proofs

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turophile

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Here's the problem:

Prove the following inequalities for 0 < x < ?/2.

(a) sec x > tan x > sin x

(b) tan x > x > sin x

I can see these inequalities hold by inspection of the graphs. For (b) at least, I think I'm supposed to use the theorem about the sign of the derivative:

Let f be continuous on an interval I and differentiable on its interior.
(i) If f'(x) > 0 throughout the interior of I, then f is an increasing function on I.
(ii) If f'(x) < 0 throughout the interior of I, then f is a decreasing function on I.

I have used this in another inequality proof:

Let g and h be continuous on an interval I and differentiable on its interior, and let a be a point of I. Prove that if g'(x) > h'(x) throughout the interior, and if g(a) ? h(a), then g(x) > h(x) for all x > a in I.

To do that proof, I applied the theorem about the sign of the derivative to the function f = g - h. However, I don't see how to get started doing this with these trig function inequality proofs. Any hints?
 
turophile said:
Here's the problem:

Prove the following inequalities for 0 < x < ?/2.

(a) sec x > tan x > sin x

sec(x) = 1/cos(x)

we know

1 > sin(x) when 0<x<?/2 ? 1/cos(x) > sin(x)/cos(x)

also

1 > cos(x) when 0<x<?/2 ? sin(x)/cos(x) > sin(x)

(b) tan x > x > sin x

I can see these inequalities hold by inspection of the graphs. For (b) at least, I think I'm supposed to use the theorem about the sign of the derivative:

Let f be continuous on an interval I and differentiable on its interior.
(i) If f'(x) > 0 throughout the interior of I, then f is an increasing function on I.
(ii) If f'(x) < 0 throughout the interior of I, then f is a decreasing function on I.

I have used this in another inequality proof:

Let g and h be continuous on an interval I and differentiable on its interior, and let a be a point of I. Prove that if g'(x) > h'(x) throughout the interior, and if g(a) ? h(a), then g(x) > h(x) for all x > a in I.

To do that proof, I applied the theorem about the sign of the derivative to the function f = g - h. However, I don't see how to get started doing this with these trig function inequality proofs. Any hints?
Click to expand...
 
Thanks! And I think I see now how to approach (b): Show that tan(0) = 0 = sin(0). Also show that on 0 < x < ?/2, tan(x) is concave up, sin(x) is concave down, x is not concave, and all three functions are increasing. Unless I'm missing something (am I?), I think that's all I need to show to prove the inequality in (b).
 
Hello, turophile!

Are we allowed a geometric proof?

\(\displaystyle \text{Prove the following inequalities for }0 < x < \tfrac{\pi}{2}\)

. . \(\displaystyle (a)\;\sec x \:>\: \tan x \:>\: \sin x\)

. . \(\displaystyle (b)\;\tan x \:>\: x \:>\: \sin x\)
Click to expand...

Part (b) is a classic inequality, used in the differentiation of \(\displaystyle \sin x.\)

Draw acute angle \(\displaystyle x\) in a unit circle.
Draw the perpendiculars shown in the diagram.

Code: 
                  C
      *           *
      |    *  A * |
      |       *   |
      |     * | * |
      |   *   |  *|
      | * x   |   |
      *-------*---*
      O       B   D

\(\displaystyle \text{We have: }\:OA \,=\, OD \,=\, 1.\)

\(\displaystyle \text{For any positive acute angle }x\text{, we have: }\:AB \:<\:\text{arc}(AD) \:<\:CD\;\;\;{\bf[1]}\)


\(\displaystyle \text{We find that:}\)

. . \(\displaystyle \begin{array}{cccccc}\sin x \:=\:\dfrac{AB}{OA} \:=\:\dfrac{AB}{1} &\Rightarrow& AB \:=\:\sin x \\ \\[-3mm] \text{arc}(AD) \:=\:r\theta \:=\:1\!\cdot\!x &\Rightarrow& \text{arc}(AD) \:=\:x \\ \\[-3mm] \tan x \:=\:\dfrac{CD}{OD} \:=\:\dfrac{CD}{1} &\Rightarrow& CD \:=\:\tan x \end{array}\)


\(\displaystyle \text{Substitute into {\bf[1]}: }\;\sin x \:<\: x \:<\:\tan x\)


\(\displaystyle \text{Therefore: }\;\tan x \:>\:x\:>\:\sin x\)

 
That's great, Soroban. That's a very clear proof to me. But the geometric principles you use are introduced in the next section of my calculus textbook.

The proof of (b) I'm supposed to do needs to leverage this proof (I think):

Let g and h be continuous on an interval I and differentiable on its interior, and let a be a point of I. Prove that if g'(x) > h'(x) throughout the interior, and if g(a) ? h(a), then g(x) > h(x) for all x > a in I. PROOF: We know that if f(x) is continuous on an interval I and differentiable on its interior, then if f'(x) > 0 throughout the interior of I, then f is an increasing function on I. Let f = g – h. Since g'(x) > h'(x) throughout the interior of I, g'(x) – h'(x) = f'(x) > 0 throughout the interior of I. Since g(a) ? h(a), g(a) – h(a) = f(a) ? 0. Hence g(x) – h(x) > 0, for x > a in I. Therefore g(x) > h(x) for all x > a in I, QED.

Now on to the proof of (b):

Let p(x) = tan x, q(x) = x, and r(x) = sin x. Then:

p'(x) = sec[sup:rc1e81n7]2[/sup:rc1e81n7]x
q'(x) = 1
r'(x) = cos x

Let I = [0, ?/2). Then p(x), q(x), and r(x) are continuous on I and differentiable on its interior.

For 0 < x < ?/2, 0 < cos x < 1, so q'(x) > r'(x). Also on that interval, 1/(cos x) = sec x > 1, so sec[sup:rc1e81n7]2[/sup:rc1e81n7]x > 1[sup:rc1e81n7]2[/sup:rc1e81n7] = 1. Thus p'(x) > q'(x). We also know that p(x) = q(x) = r(x) when x = 0. So by the earlier proof, p(x) > q(x) and q(x) > r(x) for 0 < x < ?/2. Therefore tan x > x > sin x for 0 < x < ?/2, QED.

Anyway, something along these lines is the only approach I can think of (short of your nice geometric proof).
 
turophile said:
That's great, Soroban. That's a very clear proof to me. But the geometric principles you use are introduced in the next section of my calculus textbook.
Click to expand...

That would be strange - because the famous limit theorem

\(\displaystyle \lim_{x \to 0}\left [\frac{sin(x)}{x}\right ] \ = \ 1\)

is proven using that geometric principle (at least in my school and book).
 
Here's how we did that one:

lim{x?0} (sin x)/x = lim{x?0} [sin(0+x) - sin 0]/x = sin'(0) = cos 0 = 1
 
turophile said:
Here's how we did that one:

lim{x?0} (sin x)/x = lim{x?0} [sin(0+x) - sin 0]/x = sin'(0) = cos 0 = 1
Click to expand...

So how did they prove d/dx [sin(x)] = cos(x)?
 
It went like this:

The formulas for the derivatives of sin and cos are:

d/ds (sin s) = cos, d/ds (cos s) = - sin s

Proof:

We first treat the case in which - ?/2 < s < ?/2. According to [a formula used in an earlier chapter to calculate arc length on the circle],

s = int{from 0 to y} 1/sqrt(1 - t[sup:k6773fvt]2[/sup:k6773fvt]) dt, (- 1 < y < 1).

(The formula is stated there only for y ? 0; by symmetry, it holds as well for y ? 0.) Note that this equation expresses s in terms of y = sin s. By the Fundamental Theorem of Calculus:

ds/dy = 1/sqrt(1 - y[sup:k6773fvt]2[/sup:k6773fvt]) = 1/x > 0, (- 1 < y < 1).

By the Inverse-Function Theorem [proven in an earlier chapter], dy/ds = x; that is,

d/ds (sin s) = dy/ds = x = cos s, (- ?/2 < s < ?/2).

Next, from sin[sup:k6773fvt]2[/sup:k6773fvt] s + cos[sup:k6773fvt]2[/sup:k6773fvt] s = 1, we get, differentiating implicity,

2 sin s cos s + 2 cos s d/ds (cos s) = 0.

Since cos s ? 0 we may divide. This give us

d/ds (cos s) = - sin s.

We thus get the theorem for - ?/2 < s < ?/2. To extend the results to all values of s we may proceed in steps of size ?/2, with the help of

sin s = cos(s - ?/2), cos s = - sin(s - ?/2).

Suppose we know the results for s - ?/2. Then we can obtain them for s:

d/ds (sin s) = d/ds (cos(s - ?/2)) = - sin(s - ?/2) = cos s; and similarly for d/ds (cos s).
 
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