What can we say about $a\cos^2\theta + b\sin^2\theta?$

Question

Given real numbers $a,b$ what can we say about $a\cos^2\theta + b\sin^2\theta$ and why?

My hypothesis is that the number should be between $a$ and $b$ but I can't see why.

Answer 1

Write it as $a \cdot \cos^2 \theta + a \cdot \sin^2 \theta + (b - a) \cdot \sin^2 \theta$ which gives you $a + (b - a) \cdot \sin^2 \theta$.

Answer 2

I thought that was an equation for an ellipse, but it's a bit off. But you're right about the bounds. Here is a quick proof: $$ a = a(cos^2(\theta) + sin^2(\theta)) <= a cos^2(\theta) + b sin^2(\theta) <= b(cos^2(\theta) + sin^2(\theta)) = b. $$

Answer 3

Notice that $\sin^2 \theta + \cos^2 \theta = 1$ then your expression is a weighted mean of $a$ and $b$.

Answer 4

Note that $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2\theta-1=1-2\sin^2\theta$ whence $$2\cos^2\theta = \cos 2\theta +1; 2\sin^2\theta=1-2\cos 2\theta$$

and $$a \cos^2 \theta + b\sin^2 \theta=\frac 12\left((a+b)+(a-b)\cos 2\theta\right)$$ and the $\cos 2\theta$ term is between $|a-b|$ and $-|a-b|$

Answer 5

Hint

Use the fact

min$(a,b)\le a,b\le$max$(a,b)$

Answer 6

Well $a\cos^2 \theta + b\sin^2 \theta = a(\cos^2 \theta + \sin^2 \theta) + (b-a)\sin^2 \theta = a + (b-a)\sin^2 \theta$.

And the range of $\sin^2 \theta$ is $0$ to $1$ inclusive. So the range of $(b-a)\sin^2 \theta$ is $0$ to $b-a$ inclusive.

So the range of $a + (b-a)\sin^2 \theta$ is $a$ to $a+(b-a) = b$ inclusive.

In other words if $a < b$ then

$a \le a+(b-a)\sin^2 \theta =a\cos^2 \theta + b\sin^2 \theta \le b$.

(And if $b < a$ then $b \le a\cos^2 \theta + b\sin^2 \theta \le a$.)

(And if $a=b$ then $a\cos^2 \theta + b\sin^2 \theta = a = b$.)