Prove that: sinA+sin2A+sin4A+sin5A=4cos(A2)cos(3A2)sin3A - Mathematics

प्रश्न

Prove that:
$\sin A + \sin 2 A + \sin 4 A + \sin 5 A = 4 \cos (\frac{A}{2}) \cos (\frac{3 A}{2}) \sin 3 A$

बेरीज

उत्तर

Consider LHS:
$\sin A + \sin 2 A + \sin 4 A + \sin 5 A$
$= 2 \sin (\frac{A + 2 A}{2}) \cos (\frac{A - 2 A}{2}) + 2 \sin (\frac{4 A + 5 A}{2}) \cos (\frac{4 A - 5 A}{2}) {∵ \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})}$
$= 2 \sin (\frac{3}{2} A) \cos (- \frac{A}{2}) + 2 \sin (\frac{9}{2} A) \cos (- \frac{A}{2})$
$= 2 \sin (\frac{3}{2} A) \cos (\frac{A}{2}) + 2 \sin (\frac{9}{2} A) \cos (\frac{A}{2})$
$= 2 \cos (\frac{A}{2}) {\sin \frac{3}{2} A + \sin \frac{9}{2} A}$
$= 2 \cos (\frac{A}{2}) \times 2 \sin (\frac{\frac{3}{2} A + \frac{9}{2} A}{2}) \cos (\frac{\frac{3}{2} A - \frac{9}{2} A}{2})$
$= 4 \cos (\frac{A}{2}) \sin 3 A \cos (- \frac{3}{2} A)$
$= 4 \cos \frac{A}{2} \cos (\frac{3 A}{2}) \sin 3 A$
= RHS
Hence, LHS = RHS

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Transformation Formulae

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 6.3 Q 6.2 Q 6.4

पाठ 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

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आरडी शर्मा Mathematics [English] Class 11

पाठ 8 Transformation formulae
Exercise 8.2 | Q 6.3 | पृष्ठ १८

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Prove that: sinA+sin2A+sin4A+sin5A=4cos(A2)cos(3A2)sin3A - Mathematics

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Prove that: sinA+sin2A+sin4A+sin5A=4cos(A2)cos(3A2)sin3A - Mathematics

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