angle between polynomials let $v$ be the space of polynomials less than or equal to three and let

$$\langle p,q\rangle =  p(0)q(0)+p'(0)q'(0)+p(1)q(1)+p'(1)q'(1)$$

What is the angle between the polynomials $2x^3-3x^2$ and $1$?
I was thinking $3\pi/4$ or $4\pi/3$. I was not sure which one it was. I tried drawing a picture and thought it might be $\pi/3$ too. Can someone help me please? I am a self learner and worked on this for days. I came to exhaustion. (real analysis)
 A: It depends on how you defined angle. If it is in the standard way we define in vector spaces with inner product. Let $V$ be an $\mathbb{R}$ or $\mathbb{C}$ vector space with inner product $\langle,\rangle$ the angle $\angle(u,v)$ between vectors $u,v\in V$ is defined by
$$\angle u,v=\cos^{-1}\left(\dfrac{\langle u,v\rangle}{|u||v|}\right)$$
which is well defined because $|\langle u,v\rangle|\leq |u||v|$. In that case, assuming you are dealing with the space of polynomials $P_3(\mathbb{R})$ with the usual inner product we will proceed just computing this thing. Recall that we define
$$\langle f,g\rangle=\int_0^1f(t)g(t)dt,$$
hence, for $f(x)=2x^3-3x^2$ and $g(x)=1$ we have
$$\langle f,g\rangle = \int_0^1 2t^3-3t^2dt = -\dfrac{1}{2},$$
$$\langle f,f\rangle =\int_0^1 4t^6-12t^5+9t^4dt = \dfrac{13}{35}\Longrightarrow |f|=\sqrt{13}/\sqrt{35},$$
$$\langle g,g\rangle = \int_0^1 dt=1\Longrightarrow |g|=1.$$
Because of that
$$\angle (f,g)=\cos^{-1}\left(-\sqrt{\dfrac{13}{140}}\right)\approx1,88 \ \text{rad}$$
A: Set $p(x) = 2x^3 - 3x^2$ and $q(x) = 1$. Note that $q'(x) = 0$, so we can compute
$$\langle p, q\rangle = p(0) + p(1) = -1$$
Now we use that
$$\cos\theta = \frac{\langle p, q \rangle}{\sqrt{\langle p, p \rangle \langle q, q \rangle}}$$
Now $\langle q, q\rangle = 2$, and $$\langle p, p\rangle = p(0)^2 + p'(0)^2 + p(1)^2 + p'(1)^2$$
Using the fact that $p'(x) = 6x^2 - 6x$, we see that $p'(0) = p'(1) = 0$, so that
$$\langle p, p\rangle = (-1)^2 = 1$$
Thus $$\cos{\theta} = -\frac{1}{\sqrt{2}}$$
The solutions are $\theta = 2n\pi - \frac{3\pi}{4}$.
A: Let $p_1(x) = 2x^3-3x^2$, $p_2(x) = 1$. Let $\phi: \mathbb{P}^3 \to \mathbb{R}^4$ be given by $\phi(p) = (p(0), p'(0), p(1), p'(1))$. Then
$\langle p, q \rangle_{\mathbb{P}^3} = \langle \phi(p), \phi(q) \rangle_{\mathbb{R}^4}$, so we see that it is indeed an inner product.
We have $\phi(p_1) = (0,0,-1,0)$, $\phi(p_2) = (1, 0,1,0)$, so we have $\|p_1\|_{\mathbb{P}^3} = 1, \|p_2\|_{\mathbb{P}^3} = \sqrt{2}$, and $\langle p_1, p_2 \rangle_{\mathbb{P}^3} = \langle (0,0,-1,0),  (1,0,1,0) \rangle_{\mathbb{R}^4} = -1$.
The angle is usually defined by $\cos \theta = \frac{\langle p_1, p_2 \rangle_{\mathbb{P}^3}}{\|p_1\|_{\mathbb{P}^3} \|p_2\|_{\mathbb{P}^3}} = -\frac{1}{\sqrt{2}}$. Some solutions are $\pi \pm \frac{\pi}{4}$.
