1.

Prove that: `sin5A=5sinA-20sin^(3)A+16sin^(5)A`

Answer» LHS `=sin5A=sin(3A+2A)`
`=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(4cos^(3)A-3cosA)2sinAcosA`
`=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(4cos^(2)A-3)2sinAcos^(2)A`
`=(3sinA-4sin^(3)A)(1-2sin^(2)A)+[4(1-sin^(2)A)-3]2sina(1-sin^(2)A`
`=(3sinA-4sin^(3)A)(1-2sin^(2)A)+(2sinA-2sin^(3)A)(1-4sin^(2)A)`
`=3sinA-6sin^(3)A-4sin^(3)A+8sin^(5)A +2sinA-8sin^(3)A-2sin^(3)A+8sin^(5)A`
`5sinA-20sin^(3)A+16sin^(5)A` = RHS Hence Proved.


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