# Trigonometric inequality

#### MarkFL

Staff member
Show that :

$$\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2$$

#### Albert

##### Well-known member
Show that :

$$\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2$$
left side=
$$\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}$$
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
Are you sure , right side is correct ?

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#### MarkFL

Staff member
left side=
$$\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}$$
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
Are you sure , right side is correct ?
Yes, it is correct...it appears you are assuming the two sinusoidal factors are in phase with one another, that is for $a=b$. In this case, then your result is equivalent to that which I gave.

#### Albert

##### Well-known member
if it is correct then ,we must prove
$\dfrac {a^2+b^2}{2}\leq (\dfrac{a+b}{2})^2=\dfrac {a^2+b^2}{4}+{ab}$
for all $a,b \in R$
${\therefore \dfrac {a^2+b^2}{4}\leq ab}$
how about if ab<0,then it does not fit

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#### MarkFL

Staff member
if it is correct then ,we must prove
$\dfrac {a^2+b^2}{2}\leq (\dfrac{a+b}{2})^2=\dfrac {a^2+b^2}{4}+{ab}$
for all $a,b \in R$
${\therefore \dfrac {a^2+b^2}{4}\leq ab}$
how about if ab<0,then it does not fit
It appears you are on the right track here, but have made some algebraic errors.

#### Albert

##### Well-known member
sorry ,I have made some algebraic errors I will try to use another approach

#### MarkFL

Staff member
sorry ,I have made some algebraic errors I will try to use another approach
Your errors are quite minor, and in fact leads to a much simpler approach than I have. #### Albert

##### Well-known member
I think ,I should take a rest ,and have a cup of tea or coffee

#### Albert

##### Well-known member
for some x,and a,b if left side $\leq 0$
then it holds naturely
now we assume both sides are positive
if a=b then the original inequality holds
if a>b then :$1+b^2\leq left \,\, side \leq 1+a^2$
$1+b^2\leq right \,\, side \leq 1+a^2$
if a<b then :$1+a^2\leq left \,\, side \leq 1+b^2$
$1+a^2\leq right \,\, side \leq 1+b^2$

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#### anemone

##### MHB POTW Director
Staff member
My solution:

I first expand the LHS of the inequality and get:

$$\displaystyle ( {\sin x + a\cos x} )( {\sin x + b\cos x})=\sin^2 x+(a+b)\sin x \cos x+ab\cos^2 x$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(1-\cos^2 x)+(a+b)\sin x \cos x+ab\cos^2 x$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{ab+1}{2}+\left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x$$

Next, by applying the Cauchy-Schwarz Inequality to the part $$\displaystyle \left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x$$ yields

$$\displaystyle \left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x\le\sqrt{\left(\frac{a+b}{2}\right)^2+\left( \frac{ab-1}{2}\right)^2}\cdot\sqrt{\sin^2 2x+\cos^2 2x}$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\le \sqrt{\left(\frac{a^2b^2+a^2+b^2+1}{4}\right)}$$

$$\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\le \frac{\sqrt{(1+a^2)(1+b^2)}}{2}$$

Also, AM-GM inequality tells us that

$$\displaystyle \frac{(1+a^2)+(1+b^2)}{2}\ge\sqrt{(1+a^2)(1+b^2)}$$ or

$$\displaystyle \frac{(1+a^2)+(1+b^2)}{4}\ge\frac{\sqrt{(1+a^2)(1+b^2)}}{2}$$

$$\displaystyle \frac{2+a^2+b^2}{4}\ge\frac{\sqrt{(1+a^2)(1+b^2)}}{2}$$

Finally, by combining all that we found in the above steps, we can now conclude that

$$\displaystyle ( {\sin x + a\cos x} )( {\sin x + b\cos x})$$

$$\displaystyle =\frac{ab+1}{2}+\left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x$$

$$\displaystyle \le \frac{2+a^2+b^2}{4}+\frac{ab+1}{2}$$

$$\displaystyle \le \frac{2+a^2+b^2+2ab+2}{4}$$

$$\displaystyle \le \frac{4+a^2+b^2+2ab}{4}$$

$$\displaystyle \le 1+\frac{a^2+b^2+2ab}{4}$$

$$\displaystyle \le 1+\frac{(a+b)^2}{4}$$

$$\displaystyle \le 1+(\frac{a+b}{2})^2$$ (Q.E.D.)

#### MarkFL

Staff member
This is my proof:

Let:

$$\displaystyle A=\tan^{\small{-1}}(a)$$

$$\displaystyle B=\tan^{\small{-1}}(b)$$

Using a linear combination, we may write the inequality as:

$$\displaystyle \sqrt{(1+a^2)(1+b^2)}\sin(x+A)\sin(x+B)\le1+\left(\frac{a+b}{2} \right)^2$$

Let:

$$\displaystyle f(x)=\sin(x+A)\sin(x+B)$$

Thus:

$$\displaystyle f'(x)=\sin(2x+A+B)$$

$$\displaystyle f''(x)=2\cos(2x+A+B)$$

Then $f(x)$ has its maxima for:

$$\displaystyle x=\frac{(2k+1)\pi-(A+B)}{2}$$ where $$\displaystyle k\in\mathbb Z$$

We then find:

$$\displaystyle f\left(\frac{(2k+1)\pi-(A+B)}{2} \right)=\sin\left(\frac{(2k+1)\pi-(A+B)}{2}+A \right)\sin\left(\frac{(2k+1)\pi-(A+B)}{2}+B \right)=$$

$$\displaystyle \sin\left(\frac{(2k+1)\pi+A-B}{2} \right)\sin\left(\frac{(2k+1)\pi-A+B}{2} \right)=$$

$$\displaystyle \frac{\cos(A-B)-\cos((2k+1)\pi)}{2}=\frac{\cos(A-B)+1}{2}=$$

$$\displaystyle \frac{\cos(A)\cos(B)+\sin(A)\sin(B)+1}{2}=$$

$$\displaystyle \frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2\sqrt{(1+a^2)(1+b^2)}}$$

Now, we need only show:

$$\displaystyle \sqrt{(1+a^2)(1+b^2)}f\left(\frac{(2k+1)\pi-(A+B)}{2} \right)\le1+\left(\frac{a+b}{2} \right)^2$$

$$\displaystyle \frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2}\le1+\left( \frac{a+b}{2} \right)^2$$

$$\displaystyle 2+2ab+2\sqrt{(1+a^2)(1+b^2)}\le4+a^2+2ab+b^2$$

$$\displaystyle 2\sqrt{(1+a^2)(1+b^2)}\le2+a^2+b^2$$

$$\displaystyle 4a^2b^2+4a^2+4b^2+4\le a^4+2a^2b^2+4a^2+b^4+4b^2+4$$

$$\displaystyle 2a^2b^2\le a^4+b^4$$

$$\displaystyle 0\le(a^2-b^2)^2$$