# [SOLVED] Prove the trigonometric inequality?

• June 26th 2008, 08:32 AM
fardeen_gen
[SOLVED] Prove the trigonometric inequality?
Prove the inequality (sin x - 1)/(sin x - 2) + (1/2) ≥ (2 - sin x)/(3 - sin x)?
• June 26th 2008, 11:44 AM
flyingsquirrel
Hello
Quote:

Originally Posted by fardeen_gen
Prove the inequality (sin x - 1)/(sin x - 2) + (1/2) ≥ (2 - sin x)/(3 - sin x)?

$
\frac{\sin x-1}{\sin x -2}+\frac{1}{2}\geq \frac{2-\sin x}{3-\sin x} \,\,\,\,(1)$

Let's rewrite this inequality so that it is easy to work with it. Multiply by the negative term $2\underbrace{(\sin x-2)}_{\leq 0} \underbrace{(3-\sin x)}_{\geq 0}$ :

$
(1)\,: \,\frac{2(\sin x-1)(\sin x-2)(3-\sin x)}{\sin x -2}+\frac{1}{2}\cdot 2(\sin x-2)(3-\sin x)$
$\leq \frac{2(2-\sin x)(\sin x-2)(3-\sin x)}{3-\sin x}$

Simplify

$
(1)\,: \,2(\sin x-1)(3-\sin x)+(\sin x-2)(3-\sin x) \leq 2(2-\sin x)(\sin x-2)$

Subtract $2(2-\sin x)(\sin x-2)$

$
(1)\,: \,2(\sin x-1)(3-\sin x)+(\sin x-2)(3-\sin x) - 2(2-\sin x)(\sin x-2) \leq 0$

The LHS is a quadratic in terms of $\sin x$. To show that the inequality is true you may try to write it under the form $a\sin^2x+b\sin x+c \leq 0 \Longleftrightarrow aX^2+bX+c\leq 0$ with $X=\sin x$ and to compute its roots $X_1$ and $X_2$. Then, knowing the value of the roots and the sign of $a$, one can tell where is the polynomial positive/negative. Good luck ! :D
• June 27th 2008, 08:44 PM
fardeen_gen
I have a doubt...How does the sign of the inequality change in step 1 after multiplying LHS and RHS?
• June 27th 2008, 08:45 PM
fardeen_gen
Is it because the term by which we multiply is a negative term?
• June 27th 2008, 09:09 PM
Reckoner
Quote:

Originally Posted by fardeen_gen
Is it because the term by which we multiply is a negative term?

Yes. One of the basic rules when dealing with inequalities is that you can multiply or divide both sides by the same quantity, provided that you flip the sign when that quantity is negative. That is,

$\forall a,b,c\in\mathbb{R}$

$c > 0\text{ and }a < b\Rightarrow ac < bc$

$c < 0\text{ and }a < b\Rightarrow ac > bc$

And similarly for $>,\;\leq,\;\geq$.

And you can be sure that $(\sin x - 2) < 0$ and that $(3 - \sin x) > 0$ because $-1\leq\sin x\leq1\;\;\forall x\in\mathbb{R}$.

Then, as flyingsquirrel said, you have a quadratic in $\sin x$. Since you are dealing with a continuous function, all you need to do is find the roots and test the intervals between them to see which are nonpositive (every interval should be, if the inequality is to work for all $x$). Good luck!
• June 27th 2008, 11:27 PM
ticbol
Quote:

Originally Posted by fardeen_gen
Prove the inequality (sin x - 1)/(sin x - 2) + (1/2) ≥ (2 - sin x)/(3 - sin x)?

Here is one way. May not be by the book because I forgot the reasonings by the book. It is by common sense, more or less.

So we just plod along, doing the algebra of it until we have all the terms in the LHS.

(sinX -1)/(sinX -2) +1/2 >= (2 -sinX)/(3 -sinX) --------(i)

We rearrange the RHS of (i) so the terms there will be the form of those in the LFS. We multiply the RHS by (-1)/(-1)....this will not change the inequality sign,

(sinX -1)/(sinX -2) +1/2 >= (sinX -2)/(sinX -3) --------(ii)

Simplifying the LFS, common denominator = 2(sinX -2),
[2(sinX -1) +(sinX -2)] /[2(sinX -2)] >= (sinX -2)/(sinX -3)
[2sinX -2 +sinX -2] /[2sinX -4] >= (sinX -2)/(sinX -3)
(3sinX -4)/(2sinX -4) >= (sinX -2)/(sinX -3)

Cross multiply. Since both (2sinX -4) and (siX -3) are negative, because the maximum value of the sine of any angle is 1.0 only, then after cross multiplication, the inequality sign of (ii) will not change. ((If we do it long hand, we multiply both sides first by (2sinX -4), a negative, so the inequality sign will change. Then we multiply that by (sinX -3), a negative again, so the inequality will change again. It will go back actually to the original sense.))

So,
(3sinX -4)(sinX -3) >= (sinX -2)(2sinX -4)
3sin^2(X) -13sinX +12 >= 2sin^2(X) -8sinX +8
Putting them all in the LHS,
sin^2(X) -5sinX +4 >= 0 --------------------------(iii)

Here, we can analyze that, (iii), by common sense even if we don't have to solve for the roots of (iii). But let's go with the motion more,

Factoring (iii),
(sinX -4)(sinX -1) = 0

sinX -4 = 0
sinX = 4 ------------cannot be.

sinX -1 = 0
sinX = 1
X = arcsinn(1) = 90 degrees, or pi/2 radians.
Meaning, when X = pi/2, the inequality (iii) is true. The LHS = 0.

No more roots.

By the book, we'd try the interval [0,pi/2)?
Then the intervals (pi/2,pi], (pi,3pi/2] and (3pi/2,2pi)?

We do that and we will find that in any of those 4 intervals, all values of X will result into Inequality (iii) being true. Indirectly, the original Inequality (i) is true also.

That proves it.

--------

If sinX were negative, the -5sinX in (iii) will become positive, and the sin^2(X) will be positive also, hence the LFS will always be greater than 0.

If sinX were positive, say, 0.9999, in (iii),
(0.9999)^2 -5(0.9999) +4 >=? 0
0.00030001 >=? 0
Yes, so, OK.

If sinX were positive, say, 0.57, in (iii),
(0.57)^2 -5(0.57) +4 >=? 0
1.4749 >=? 0
Yes, of course.

If sinX = 0.0008,
(0.0008)^2 -5(0.0008) +4 >=? 0
3.996 >=? 0
You bet.