prove that
rn
sin 20 sin 40 sin 80 =under root 3/8
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Answered by
166
sin(20)sin(40) = (1/2)[cos(20) - cos(60)]
=sin(20)sin(40)sin(80)
= (1/2)[cos(20) - cos(60)]sin(80)
= (1/2)[cos(20)sin(80) - cos(60)sin(80)]
= (1/2)[(1/2)(sin(100)+sin(60)) - (1/2)(sin(140)+sin(20))]
= (1/4)[sin(80) + sin(60) - sin(40) - sin(20)]
= (1/4)sin(60), since sin(80) - sin(40) - sin(20) = sin(60+20) - sin(60-20) - sin(20) = 2cos(60)sin(20) - sin(20) = 0
= √3/8
=sin(20)sin(40)sin(80)
= (1/2)[cos(20) - cos(60)]sin(80)
= (1/2)[cos(20)sin(80) - cos(60)sin(80)]
= (1/2)[(1/2)(sin(100)+sin(60)) - (1/2)(sin(140)+sin(20))]
= (1/4)[sin(80) + sin(60) - sin(40) - sin(20)]
= (1/4)sin(60), since sin(80) - sin(40) - sin(20) = sin(60+20) - sin(60-20) - sin(20) = 2cos(60)sin(20) - sin(20) = 0
= √3/8
arjunash93own123:
Which formula Ur added
Answered by
151
sin(20)sin(40) = (1/2)[cos(20) - cos(60)]
=sin(20)sin(40)sin(80)
= (1/2)[cos(20) - cos(60)]sin(80)
= (1/2)[cos(20)sin(80) - cos(60)sin(80)]
= (1/2)[(1/2)(sin(100)+sin(60)) - (1/2)(sin(140)+sin(20))]
= (1/4)[sin(80) + sin(60) - sin(40) - sin(20)]
= (1/4)sin(60), since sin(80) - sin(40) - sin(20) = sin(60+20) - sin(60-20) - sin(20) = 2cos(60)sin(20) - sin(20) = 0
= √3/8
=sin(20)sin(40)sin(80)
= (1/2)[cos(20) - cos(60)]sin(80)
= (1/2)[cos(20)sin(80) - cos(60)sin(80)]
= (1/2)[(1/2)(sin(100)+sin(60)) - (1/2)(sin(140)+sin(20))]
= (1/4)[sin(80) + sin(60) - sin(40) - sin(20)]
= (1/4)sin(60), since sin(80) - sin(40) - sin(20) = sin(60+20) - sin(60-20) - sin(20) = 2cos(60)sin(20) - sin(20) = 0
= √3/8
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