If \( \sin 40^{\circ}=p \), write the ff in termes \( p \) \( 2 \sin 220=p^{2} \)

To express \( 2 \sin 220^\circ \) in terms of \( p \), we first find \( \sin 220^\circ \). We can break this down as: \[ \sin 220^\circ = \sin (180^\circ + 40^\circ) = -\sin 40^\circ = -p \] Now substituting this back into the equation, we have: \[ 2 \sin 220^\circ = 2(-p) = -2p \] Therefore, \( 2 \sin 220^\circ = -2p \). Now let's look at the second part of your question. We need to compare this result with \( p^2 \): \[ p^2 = p \cdot p \] Thus, we have \[ -2p = p^2. \] Now, we can rearrange this to find a polynomial: \[ p^2 + 2p = 0. \] To solve for \( p \), we can factor out \( p \): \[ p(p + 2) = 0. \] This gives two results: \( p = 0 \) or \( p + 2 = 0 \), meaning \( p = -2 \). However, since \( p \) represents \( \sin 40^\circ \), we conclude that \( p = \sin 40^\circ \) is the relevant solution here.