If A=45° verify that sin2A=2sinAcosA
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Answered by
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★ TRIGONOMETRIC REDUCTIONS ★
A = 45°
sin 2A = sin 90° = 1
sin45° ( cos45° ) = 1/ √2 ( 1/ √2 ) = 1/ 2
∴ 2sinAcosA = 1
HENCE VERIFIED
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A = 45°
sin 2A = sin 90° = 1
sin45° ( cos45° ) = 1/ √2 ( 1/ √2 ) = 1/ 2
∴ 2sinAcosA = 1
HENCE VERIFIED
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Answered by
50
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