(i) Find an expression for \( \sin 140^\circ \).

Use the angle identity: \( \sin (180^\circ – \theta) = \sin \theta \).

\[ \sin 140^\circ = \sin (180^\circ – 40^\circ) = \sin 40^\circ = p \]

Answer: \( \sin 140^\circ = p \).

(ii) Find an expression for \( \cos 70^\circ \).

Use the angle identity: \( \cos (180^\circ – \theta) = -\cos \theta \).

\[ \cos 70^\circ = \cos (180^\circ – 110^\circ) = -\cos 110^\circ = -q \]

Answer: \( \cos 70^\circ = -q \).

Find an expression for \( \cos 140^\circ \).

Method 1: Use the Pythagorean identity.

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

For \( \theta = 140^\circ \), and using \( \sin 140^\circ = p \):

\[ \sin^2 140^\circ + \cos^2 140^\circ = 1 \]

\[ p^2 + \cos^2 140^\circ = 1 \]

\[ \cos^2 140^\circ = 1 – p^2 \]

\[ \cos 140^\circ = \pm \sqrt{1 – p^2} \]

Since \( 140^\circ \) is in the second quadrant, where cosine is negative:

\[ \cos 140^\circ = -\sqrt{1 – p^2} \]

Method 2: Use the double-angle formula for cosine.

\[ \cos 140^\circ = \cos (2 \cdot 70^\circ) = 2 \cos^2 70^\circ – 1 \]

From part (a)(ii), \( \cos 70^\circ = -q \):

\[ \cos 140^\circ = 2 (-q)^2 – 1 = 2q^2 – 1 \]

Answer: \( \cos 140^\circ = -\sqrt{1 – p^2} \) or \( \cos 140^\circ = 2q^2 – 1 \).

Find an expression for \( \tan 140^\circ \).

Use the definition of tangent: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).

From part (a)(i), \( \sin 140^\circ = p \).

From part (b), \( \cos 140^\circ = -\sqrt{1 – p^2} \) (Method 1) or \( \cos 140^\circ = 2q^2 – 1 \) (Method 2).

Method 1:

\[ \tan 140^\circ = \frac{\sin 140^\circ}{\cos 140^\circ} = \frac{p}{-\sqrt{1 – p^2}} = -\frac{p}{\sqrt{1 – p^2}} \]

Method 2:

\[ \tan 140^\circ = \frac{\sin 140^\circ}{\cos 140^\circ} = \frac{p}{2q^2 – 1} \]

Answer: \( \tan 140^\circ = -\frac{p}{\sqrt{1 – p^2}} \) or \( \tan 140^\circ = \frac{p}{2q^2 – 1} \).