Question: Prove the following trigonometric identity: sin5A = 16 sin⁵A - 20sin³A +5 sinA


Solution:

LHS
= sin 5A
= sin (2A +3A)
[ sin(A+B) = sinA.cosB + cosA.sinB]
= sin 2A . cos 3A + cos 2A . sin 3A

= (2sinA cosA).(4cos³A - 3cosA) + (1 -2sin²A).(3sinA - 4sin³A)

= 8sinA cos⁴A - 6sinAcos²A + 3sinA - 4sin³A - 6sin³A +8sin⁵A

= 8sin⁵A + 8sinA.cos⁴A -10sin³A - 6sinAcos²A -10sin³A +3sinA

[ cos²A = 1 - sin²A ]

= 8sin⁵A + 8sinA .cos²A . cos²A - 6sinA(1-sin²A) -10sin³A + 3sinA

= 8sin⁵A + 8sinA (1 -sin²A) (1 -sin²A) -6sinA +6sin³A - 10sin³A +3sinA

= 8sin⁵A + 8sinA (1 -sin²A)² - 4sin³A -3sinA

= 8sin⁵A + 8sinA (1 -2sin²A +sin⁴A) - 4sin³A - 4sin³A -3sinA

= 8sin⁵A +8sinA -16sin³A +8sin⁵A -4sin³A -3sinA

= 16sin⁵A - 20sin³A +5sinA
= RHS

Related Notes And Solutions:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

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