EXERCISE 12 1 If \( \cos S 2^{\circ}=k \), find each ratio in terms of \( A \). \( 1.1 \cos 128^{\circ} \) \( 1.2 \sin 38^{\circ} \) \( 1.3 \tan 218^{\circ} \) \( 1.4 \sin 308^{\circ} \) \( 1.6 \tan \left(-142^{\circ}\right) \) Hint: \( 1.5 \cos 232^{\prime} \) Draw: diagram. 2 If \( 2 \tan 25^{\circ}=p \), find each ratio in terms of \( p \). \( 2.2 \sin 205^{\circ} \) \( 2.1 \tan 155^{\circ} \) \( 2.4 \tan 65^{\circ} \) \( 2.3 \cos 335^{\circ} \) \( 2.6 \sin 295^{\circ} \) 3 If \( \sin 43^{\circ}=a \), expiess each ratio in terms of \( \alpha \). \( 3.1 \sin 223^{\circ} \) \( 3.2 \cos 317^{\circ} \) \( 3.3 \sin \left(-137^{\circ}\right) \) \( 3.4 \cos 133^{\circ} \)

Let's solve the exercise step by step, starting with the first part. ### Part 1: If \( \cos S 2^{\circ} = k \) #### 1.1 Find \( \cos 128^{\circ} \) Using the cosine identity: \[ \cos(180^{\circ} - x) = -\cos(x) \] we can express \( \cos 128^{\circ} \) as: \[ \cos 128^{\circ} = \cos(180^{\circ} - 52^{\circ}) = -\cos 52^{\circ} \] Now, we can relate \( \cos 52^{\circ} \) to \( k \): \[ \cos 52^{\circ} = \sin 38^{\circ} \quad \text{(since \( \sin x = \cos(90^{\circ} - x) \))} \] Thus, \[ \cos 128^{\circ} = -\sin 38^{\circ} \] #### 1.2 Find \( \sin 38^{\circ} \) Using the identity: \[ \sin(90^{\circ} - x) = \cos x \] we have: \[ \sin 38^{\circ} = \cos(90^{\circ} - 38^{\circ}) = \cos 52^{\circ} \] And since \( \cos 52^{\circ} = \sin 38^{\circ} \), we can express it in terms of \( k \): \[ \sin 38^{\circ} = \sqrt{1 - k^2} \] #### 1.3 Find \( \tan 218^{\circ} \) Using the identity: \[ \tan(180^{\circ} + x) = \tan x \] we have: \[ \tan 218^{\circ} = \tan(180^{\circ} + 38^{\circ}) = \tan 38^{\circ} \] Thus, \[ \tan 218^{\circ} = \frac{\sin 38^{\circ}}{\cos 38^{\circ} = \frac{\sqrt{1 - k^2}}{k} \] #### 1.4 Find \( \sin 308^{\circ} \) Using the identity: \[ \sin(360^{\circ} - x) = -\sin x \] we have: \[ \sin 308^{\circ} = -\sin 52^{\circ} \] And since \( \sin 52^{\circ} = \cos 38^{\circ} = k \): \[ \sin 308^{\circ} = -k \] #### 1.5 Find \( \cos 232^{\circ} \) Using the identity: \[ \cos(180^{\circ} + x) = -\cos x \] we have: \[ \cos 232^{\circ} = -\cos 52^{\circ} = -\sin 38^{\circ} = -\sqrt{1 - k^2} \] #### 1.6 Find \( \tan(-142^{\circ}) \) Using the identity: \[ \tan(-x) = -\tan x \] we have: \[ \tan(-142^{\circ}) = -\tan 142^{\circ} \] Using the identity: \[ \tan(180^{\circ} - x) = -\tan x \] we find: \[ \tan 142^{\circ} = -\tan 38^{\circ} \] Thus, \[ \tan(-142^{\circ}) = \tan 38^{\circ} = \frac{\sqrt{1 - k^2}}{k} \] ### Part 2: If \( 2 \tan 25^{\circ} = p \) #### 2.1 Find \( \tan 155^{\circ} \) Using the identity: \[ \tan(180^{\circ} - x) = -\tan x \] we have: \[ \tan 155^{\circ} = -\tan 25^{\circ} = -\frac{p}{2} \] #### 2.2 Find \( \sin 205^{\circ} \) Using the identity: \[ \sin(180^{\circ} + x) = -\sin x \] we have: \[ \sin 205^{\circ} = -\sin 25^{\circ} \] Using the identity: \[ \sin x = \tan x \cdot \cos x \] we can express it in terms of \( p \): \[ \sin 25^{\circ} = \frac{p}{2} \cdot \cos 25^{\circ} \] Thus, \[ \sin 205^{\circ} = -\frac{p}{2} \cdot \cos 25^{\circ} \] #### 2.3 Find \( \cos 335^{\circ} \) Using the identity: \[ \cos(360^{\circ} - x) = \cos x \] we have: \[ \cos 335^{\circ} = \cos 25^{\circ} \] #### 2.4 Find \( \tan 65^{\circ} \) Using the identity: \[ \tan(90^{\circ} - x) = \cot x \] we have: \[ \tan 65^{\circ} = \cot 25^{\circ} = \frac{1}{\tan 25^{\circ}} = \frac{2}{p} \] #### 2.5 Find \( \sin 295^{\circ} \) Using the identity: \[ \sin(360^{\circ} - x) = -\sin x \] we have: \[ \sin 295^{\circ} = -\sin 65^{\circ} \] Using the identity: \[ \sin 65^{\circ} = \cos 25^{\circ} \] Thus, \[ \sin 295^{\circ} = -\cos 25^{\circ} \] ### Part 3: If \( \sin 43^{\circ} = a \) #### 3.1 Find \( \sin 223^{\circ} \) Using the identity: \[ \sin(180^{\circ} + x) = -\sin x \] we have: \[ \sin 223^{\circ} = -\sin 43^{\circ} = -a \] #### 3.2 Find \( \cos 317^{\circ} \) Using the identity: \[ \cos(360^{\circ} - x) = \cos x \] we have: \[ \cos