solve sin10°sin50°sin60°sin70°?
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Hey there !
sin10°sin50°sin60°sin70°
= sin60° ( sin10°sin50°) sin70
= √3/4 ( 2 sin10sin50) sin70
cosC-cosD = 2sin( C+D)/2 sin(D-C)/2
= √3/4 ( cos40-cos60) sin70
= √3/4 ( cos40sin70) - √3/4 * 1/2 sin70
=√3/8( sin110+sin30) -√3/8sin70
= √3/8[ sin110+ 1/2 - sin70]
= √3/8*1/2+ √3/16 [ sin110-sin70]
= √3/16 + √3/16[ 2 cos90 sin20]
=√3/16
sin10°sin50°sin60°sin70°
= sin60° ( sin10°sin50°) sin70
= √3/4 ( 2 sin10sin50) sin70
cosC-cosD = 2sin( C+D)/2 sin(D-C)/2
= √3/4 ( cos40-cos60) sin70
= √3/4 ( cos40sin70) - √3/4 * 1/2 sin70
=√3/8( sin110+sin30) -√3/8sin70
= √3/8[ sin110+ 1/2 - sin70]
= √3/8*1/2+ √3/16 [ sin110-sin70]
= √3/16 + √3/16[ 2 cos90 sin20]
=√3/16
kajalpal660:
thnks
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