Express sin5x in powers of sin x
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Show that Sin 5 theta=5Sin theta -20 Sin^3 theta +16 Sin^5 theta?
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9 ANSWERS

Adrija Dalal, B. Sc Vivekananda College, Thakurpukur, Aligunj Rishi Rajnarayana Balika Vidyalaya (2020)
Answered Feb 21, 2018
Here theta is denoted by A
sin5A = sin(3A+2A)
= sin3Acos2A+cos3Asin2A
=(3sinA-4sin³A)(1–2sin²A) + cos(2A+A)sin2A
=3sinA-10sin³A+8sin^5A+ [cos2AcosA- sin2AsinA]sin2A
=3sinA-10sin³A+8sin^5A+ [(1–2sin²A)cosA- 2sin²AcosA]2sinAcosA
=3sinA-10sin³A+8sin^5A+[cosA-4sin²AcosA]2sinAcosA
=3sinA-10sin³A+8sin^5A+2sinAcos²A- 8sin³Acos²A
=3sinA-10sin³A+8sin^5A+2sinA(1-sin²A)-
8sin³A(1-sin²A)
=3sinA-10sin³A+8sin^5A+2sinA-2sin³A-8sin³A+8sin^5A
=5sinA-20sin³A+16sin^5A[Proved]
Still have a question? Ask your own!
What is your question?
9 ANSWERS

Adrija Dalal, B. Sc Vivekananda College, Thakurpukur, Aligunj Rishi Rajnarayana Balika Vidyalaya (2020)
Answered Feb 21, 2018
Here theta is denoted by A
sin5A = sin(3A+2A)
= sin3Acos2A+cos3Asin2A
=(3sinA-4sin³A)(1–2sin²A) + cos(2A+A)sin2A
=3sinA-10sin³A+8sin^5A+ [cos2AcosA- sin2AsinA]sin2A
=3sinA-10sin³A+8sin^5A+ [(1–2sin²A)cosA- 2sin²AcosA]2sinAcosA
=3sinA-10sin³A+8sin^5A+[cosA-4sin²AcosA]2sinAcosA
=3sinA-10sin³A+8sin^5A+2sinAcos²A- 8sin³Acos²A
=3sinA-10sin³A+8sin^5A+2sinA(1-sin²A)-
8sin³A(1-sin²A)
=3sinA-10sin³A+8sin^5A+2sinA-2sin³A-8sin³A+8sin^5A
=5sinA-20sin³A+16sin^5A[Proved]
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