EXERCISE (a) (1) \( \sin 236^{\circ} \cos 169^{\circ}+\cos 371^{\circ} \sin \left(-56^{\circ}\right) \)

We want to compute   sin 236° · cos 169° + cos 371° · sin (–56°). Step 1. Rewrite angles using periodicity and reference angles. • Notice that 371° is more than 360°. Since cosine is periodic with period 360°,   cos 371° = cos (371° – 360°) = cos 11°. • Also, sin (–56°) = – sin 56°  (since sine is an odd function). • For 236°, note that 236° = 180° + 56°. Recall that   sin (180° + x) = – sin x,  thus sin 236° = – sin 56°. • Next, 169° can be expressed as 169° = 180° – 11°. Since   cos (180° – x) = – cos x,  we have cos 169° = – cos 11°. Step 2. Substitute these into the original expression.   Original expression = sin 236° · cos 169° + cos 371° · sin (–56°)     = (– sin 56°)(– cos 11°) + cos 11° (– sin 56°). Step 3. Simplify term by term.   First term: (– sin 56°)(– cos 11°) = sin 56° · cos 11°.   Second term: cos 11°(– sin 56°) = – sin 56° · cos 11°. Step 4. Add the two results.   Result = sin 56° · cos 11° – sin 56° · cos 11° = 0. Thus, the evaluated expression is 0.