Question

If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, find A and B. (a) 30°, 15° (b) 45°, 60° (c) 0, 90° (d) 45°, 45°

Answer 1

Hope u like my process
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$= > \sin(a + 2b) = \frac{ \sqrt{3} }{2} \\ \\ = > \sin(a + 2b) = \sin( \frac{\pi}{3} ) \\ \\ = > \bf \: a + 2b = \frac{\pi}{3} ........(1) \\ \\ \\ = > \cos(a + 4b) = 0 \\ \\ = > \cos(a + 4b) = \cos( \frac{\pi}{2 } ) \\ \\ = > \bf \: a + 4b = \frac{\pi}{2} .......(2)$
Comparing eq (1) and (2) we get,

$= > \bf \: a + 4b = \frac{\pi}{2} \\ - \bf \: \: \: \: \: \: a + 2b = \frac{\pi}{3} \\ ................................ \\ = > \bf \: 4b - 2b = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \\ \\ = > 2b = \frac{\pi}{6} \\ \\ = > \bf \: b = \underline{\frac{\pi}{12} \: \: \: or \: \: \: 1 5°}$

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So,

$= > \bf \: a + 2b = \frac{\pi}{3} \\ \\ = > a = \frac{\pi}{3} - 2b = \frac{\pi}{3} - \frac{\pi}{6} \\ \\ = > \bf \: a = \underline{ \frac{\pi}{6} \: \: \: or \: \: \: 30° }$
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So, a = 30° and b =15 °
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