Question
EXERCISE (a) (1) \( \sin 236^{\circ} \cos 169^{\circ}+\cos 371^{\circ} \sin \left(-56^{\circ}\right) \)
Ask by Burns Reyes. in South Africa
Mar 03,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The expression simplifies to 0.
Solution
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.
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution The Deep Dive
Did you know that \( \sin \) and \( \cos \) functions can help us decipher angles in the unit circle? For example, \( \sin 236^{\circ} \) is equivalent to \( \sin (180^{\circ} + 56^{\circ}) = -\sin 56^{\circ} \) because \( 236^{\circ} \) is in the third quadrant where sine is negative. On the other hand, \( \cos 169^{\circ} \) can be identified as \( -\cos (180^{\circ} - 11^{\circ}) = -\cos 11^{\circ} \). Now, add in that \( \cos 371^{\circ} = \cos (360^{\circ} + 11^{\circ}) = \cos 11^{\circ} \) and \( \sin (-56^{\circ}) = -\sin 56^{\circ} \). When you piece it all together, you will find some neat cancellations that keep math both interesting and rewarding!
