If \( 2 \tan 25^{\circ}=p \), determine the following in terms of \( p \) \( 4.1 \sin 205^{\circ} \) \( 4.2 \cos \left(-115^{\circ}\right) \)

We are given \[ 2\tan 25^\circ = p. \] This means \[ \tan 25^\circ = \frac{p}{2}. \] Because \(\tan 25^\circ=\frac{\sin25^\circ}{\cos25^\circ}\), we have \[ \sin25^\circ = \tan25^\circ\,\cos25^\circ = \frac{p}{2}\cos25^\circ. \] To express \(\cos25^\circ\) in terms of \(p\), recall that \[ \cos25^\circ=\frac{1}{\sqrt{1+\tan^2 25^\circ}}=\frac{1}{\sqrt{1+\left(\frac{p}{2}\right)^2}}=\frac{1}{\sqrt{1+\frac{p^2}{4}}}=\frac{2}{\sqrt{4+p^2}}. \] Thus, \[ \sin25^\circ = \frac{p}{2}\cdot\frac{2}{\sqrt{4+p^2}}=\frac{p}{\sqrt{4+p^2}}. \] Now we solve the two parts. --- **1. \(4\sin205^\circ\)** Notice that \[ 205^\circ = 180^\circ + 25^\circ, \] and using the sine shift formula, \[ \sin(180^\circ+ \theta)=-\sin\theta, \] we have \[ \sin205^\circ = -\sin25^\circ. \] Thus, \[ 4\sin205^\circ = -4\sin25^\circ. \] Substitute the expression for \(\sin25^\circ\): \[ 4\sin205^\circ = -4\cdot\frac{p}{\sqrt{4+p^2}} = -\frac{4p}{\sqrt{4+p^2}}. \] --- **2. \(4.2\cos(-115^\circ)\)** Since cosine is an even function, \[ \cos(-115^\circ)=\cos115^\circ. \] Also, express \(115^\circ\) as \[ 115^\circ = 180^\circ - 65^\circ. \] Using the cosine identity, \[ \cos(180^\circ-\theta)=-\cos\theta, \] we get \[ \cos115^\circ = -\cos65^\circ. \] Now, recall that \[ \cos65^\circ = \sin(90^\circ-65^\circ)=\sin25^\circ. \] Hence, \[ \cos115^\circ = -\sin25^\circ, \] and therefore, \[ 4.2\cos(-115^\circ)=4.2\cos115^\circ= -4.2\sin25^\circ. \] Substitute the expression for \(\sin25^\circ\): \[ 4.2\cos(-115^\circ)= -4.2\cdot\frac{p}{\sqrt{4+p^2}} = -\frac{4.2p}{\sqrt{4+p^2}}. \] --- **Final Answers:** \[ \quad 4\sin205^\circ = -\frac{4p}{\sqrt{4+p^2}} \] \[ \quad 4.2\cos(-115^\circ) = -\frac{4.2p}{\sqrt{4+p^2}}. \]