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The question

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds

The idea

Let's note the sides of the triangle as $a,b,c$

This means that according to the Pythagoras $a^2=b^2+c^2$.

I also tried replacing $\sin B =\frac{AC}{BC}$ and $\sin C= \frac{AB}{BC}$, but got to nothing useful.

I thought of using the sine rule...but again, I got to nothing...

I hope one of you can help me! Thank you!

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    $\begingroup$ I took the liberty to improve formatting. $\endgroup$
    – Sahaj
    Feb 21 at 10:44
  • $\begingroup$ As the sum of $B$ and $C$ equals $\frac{\pi}{2}$, you do realise that $\sin C = \cos B$? :-) $\endgroup$
    – Dominique
    Feb 21 at 12:59

5 Answers 5

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WLOG the hypotenuse is 1. Let the legs be $a,b$. Denote $ab = x$. $$(1-a)(1-b) = \frac{a^2b^2}{1+a+b+ab} \le \frac{a^2b^2}{1+2\sqrt {ab}+ab} = \left(\frac{x}{1+\sqrt x}\right)^2$$ Now, consider: $$\frac{x}{1+\sqrt x} = \sqrt x - \frac{\sqrt x}{\sqrt x+1} = \sqrt x -1 + \frac 1{\sqrt x+1} = \sqrt x+1 + \frac 1{\sqrt x+1} -2$$ The function $f(y) = y+\frac 1y$ is increasing for $y>1$, so the expression attains max value when $\sqrt x$ is maximum. Verify that this happens when $a=b$ as: $$\sqrt{ab} \le \sqrt{(a^2+b^2)/2} = \frac 1{\sqrt 2}$$by AM-GM. Equality when $a=b$.


Claim: $\forall y>1$, $y+1/y$ is strictly increasing.
Proof: Let $m>n>1$. Observe that $$\left(m+\frac 1m\right)-\left(n+\frac 1n\right)$$ $$ = (m-n)+\left(\frac 1m - \frac 1n\right)$$ $$= (m-n)+\left(\frac {n-m}{mn}\right)$$ $$= (m-n)\left(1 - \frac 1{mn}\right) >0$$

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  • $\begingroup$ Thaank you so much for your idea. Is there any way to demonstrate it without using functions? (I didn't learned about them yet) $\endgroup$ Feb 21 at 16:46
  • $\begingroup$ @IONELABUCIU we can show the function is increasing without calculus too. Is that what you're asking? $\endgroup$
    – D S
    Feb 21 at 16:56
  • $\begingroup$ No, i mean if we can show that $\sqrt{x}+\frac{1}{\sqrt{x}}$ get its max value when $a=b$ without using functions....? $\endgroup$ Feb 21 at 16:58
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    $\begingroup$ @IONELABUCIU I say something different. I say $1+\sqrt{x} + \frac 1{1+\sqrt{x}}$ attains its max value when $\sqrt{x}$ is max. Then I say that this happens when $a=b$. $\endgroup$
    – D S
    Feb 21 at 17:02
  • $\begingroup$ @IONELABUCIU where do you get so many questions from? $\endgroup$
    – D S
    Feb 21 at 17:30
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As it was pointed out in other answers and comments, this amounts to determining the maximum value of $f(x)=(1-\sin x)(1-\cos x)$ for $x \in [0, \frac{\pi}{2}]$. Since $f$ is differentiable, the maximum will occur at the extreme points of the interval or at a point in the interior where the derivative is zero. Now,

$$ f(x)=(1-\cos x)(1-\sin x) =\frac 12 (\sin x + \cos x - 1)^2, $$

which makes the zeros of $f'$ easy to compute and the result follows.

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Hint : It is a right angle triangle so, $B + C = 90^{\circ}$.

So, this can be re-written as $$(1-\sin B)(1-\cos B)$$

I hope you will be proceed from to get maxima at $B=45^{\circ}$ and then the final result

Edit : Full solution added There are several ways to do this... I'll use the method (Let x=B for simplicity)

$$(1-\sin x)(1-\cos x)=1- \sin x - \cos x + \sin x \cos x =1- \sin x - \cos x+\frac{\sin 2x}{2}$$

Now differentiate and equate to zero, $$- \cos x + \sin x+\cos 2x=0$$ $$\Rightarrow \cos 2x = \cos x - \sin x $$ $$\Rightarrow \cos^2x-\sin^2x =\cos x - \sin x $$ $$\Rightarrow (\cos x - \sin x)(\cos x + \sin x)=(\cos x - \sin x) $$ $$\Rightarrow (\cos x - \sin x)(\cos x + \sin x - 1)=0$$

So, we have either $\sin x = \cos x$ or $\sin x + \cos x = 1$. Remember the $x \in (0,\frac{\pi}{2})$ So, first equation gives $x = \frac{\pi}{4}$ and second equation gives $x = 0 $ or $ \frac{\pi}{2}$.

Now, putting in the values we realise tht $x = 0$ and $x = \frac{\pi}{2}$ are points of minima and give the value $0$ while $x=\frac{\pi}{4}$ is the point of maxima.

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  • $\begingroup$ Thank you for your answer. May i ask why when B=45 the value is minimal and how does this help us to get the final result...its one of my first doing a trigonometry problem... $\endgroup$ Feb 21 at 10:43
  • $\begingroup$ I took the liberty to inprove MathJax on the post. $\endgroup$
    – Sahaj
    Feb 21 at 10:45
  • $\begingroup$ @IONELABUCIU There are many ways to achieve the result from here. You may try to differentiate or maybe use AM-GM(this might be hard... if its ur first time) or by intuition we can say the maxima occurs when both terms are equal in this $\endgroup$ Feb 21 at 11:24
  • $\begingroup$ @Sahaj Appreciate it. Thank You $\endgroup$ Feb 21 at 11:25
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    $\begingroup$ @IONELABUCIU I have added the full solution. PLease check it out and lmk if you understood $\endgroup$ Feb 21 at 15:15
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$\sin B = \dfrac{b}{a} , \sin C = \dfrac{c}{a} $, so

$ (1 - \sin B) (1 - \sin C) = \dfrac{(a - b)(a - c)}{a^2}$

We can take $a = 1$ without loss of generality because we talking about angles here not lengths. Therefore, with $a = 1$ the expressions becomes

$ (1 - \sin B) (1 - \sin C) = (1 - b)(1 - c) $

And we want to maximize this function subject to $b^2 + c^2 = 1 $

Using Lagrange multiplier method,

$ g( b, c, \lambda) = (1 - b)(1 - c) + \lambda( b^2 + c^2 - 1 ) $

Taking the two partial derivatives of $g$ with respect to $b$ and $c$,

$g_b = - (1 - c) + \lambda (2 b) = 0 $

$g_c = - (1 - b) + \lambda (2 c) = 0 $

Therefore, at the maximum we have

$ \lambda = \dfrac{(1 - c)}{2 b} = \dfrac{(1 - b)}{2 c} $

i.e.

$ c (1 - c) = b (1 - b) $

So that

$ b^2 - c^2 + c - b = 0 $

which is a hyperbola in the $(b,c)$ plane. We have to intersect this hyperbola with the circle

$ b^2 + c^2 = 1 $

When you do, you get the following solutions:

$(b, c) = (0,1), (1, 0), (- \dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}) , (\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}) $

The only valid solution is $(b, c) = (\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}) $

At this point, the value of the function is

$ (1 - b )(1 - c) = ( 1 - \dfrac{1}{\sqrt{2}})^2 = \dfrac{(\sqrt{2}- 1)^2 }{2} $

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In a triangle $\triangle ABC$ $$\cos A+\cos B+\cos C=1+4\sin \tfrac A2\sin\tfrac B2\sin\tfrac C2.$$ Source. Sign mistake.

When $A=90^\circ,$ $$\sin\tfrac B2\sin\tfrac C2=\frac{\cos B+\cos C-1}{2\sqrt2}$$ $$4\sin^2\tfrac B2\sin^2\tfrac C2=\frac{(\cos B+\cos C-1)^2}{2}$$ $$(1-\sin B)(1-\sin C)=\frac{(\sqrt2\sin(B+45)-1)^2}{2}\\ \leq\frac{(\sqrt2-1)^2}2$$

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