1. ## Cauchy-Schwarz Inequality

I am supposed to prove that for all real numbers $a,b, \theta$ that
$(acos\theta+bsin\theta)^2\leq a^2+b^2$
I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

$(acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2$

Am I on the right track?

2. $=+$

$=+2+=.....$

Also, I am assuming inner product is defined as such:

$\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)dx$ because if this is the case, everything works out just fine.

3. It does not specify the inner product. I have posted all the information that was given to me.

4. Originally Posted by Jukodan
It does not specify the inner product. I have posted all the information that was given to me.
Try finishing the problem with that inner product in mind and you will have your problem completed.

5. Originally Posted by Jukodan
I am supposed to prove that for all real numbers $a,b, \theta$ that
$(acos\theta+bsin\theta)^2\leq a^2+b^2$
$\left( {a\sin (t) - b\cos (t)} \right)^2 = a^2 \sin ^2 (t) - 2ab\sin (t)\cos (t) + b^2 \cos (t)$ Note the minus.
So we get $\left( {a\cos (t) + b\sin (t)} \right)^2 = a^2 + b^2 - \left( {a\sin (t) - b\cos (t)} \right)^2$
Notice that $a^2 + b^2 - \left( {a\sin (t) - b\cos (t)} \right)^2 \le a^2+b^2$

6. $=+$

$=+2+=.....$

Spoiler:
$a^2+2ab+b^2$

$\frac{1}{\pi}\int_{-\pi}^{\pi}(a^2cos^2x)dx=a^2$

$\frac{1}{\pi}\int_{-\pi}^{\pi}(2ab*cosx*sinx)dx=0$

$\frac{1}{\pi}\int_{-\pi}^{\pi}(b^2sin^2x)dx=b^2$

$a^2+2ab+b^2$

$a^2+b^2$

$\Rightarrow (acosx+bsinx)^2\leq a^2+b^2$

7. Ah-ha!

8. Originally Posted by Jukodan
I am supposed to prove that for all real numbers $a,b, \theta$ that
$(acos\theta+bsin\theta)^2\leq a^2+b^2$
I've only gotten up to this point which is where I am not sure as to how to appply the cauchy schwarz inequality,

$(acos\theta+bsin\theta)^2=a^2+b^2-(asin\theta+bcos\theta)^2$

Am I on the right track?
The C–S inequality says that $x_1y_1+x_2y_2\leqslant (x_1^2+x_2^2)^{1/2}(y_1^2+y_2^2)^{1/2}$. Apply that with $x_1=a$, $x_2=b$, $y_1=\cos\theta$, $y_2 = \sin\theta$. Then square both sides.