# Chapter 5 - Inner Product Spaces - 5.2 Inner Product Spaces - 5.2 Exercises - Page 246: 50

$$\cos \theta=\frac{-1}{\sqrt 6}.$$

#### Work Step by Step

Let $p(x)=1+x^2,\quad q(x)=x-x^2, \quad \langle u, v\rangle=a_0b_0+2a_1b_1+a_2b_2$. The angle $\theta$ between $u$ and $v$ is given by the formula $$\cos \theta=\frac{\langle u, v\rangle}{\|u\| \cdot\|v\|}=\frac{0+0-1}{\sqrt{ 1+0+1}\sqrt{ 0+2+1}}=\frac{-1}{\sqrt 6}.$$

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