8sin20°sin40°sin80°=√3 prove that
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LHS = 8sin20° .sin40°.sin80°
=4{ 2sin20.sin80°}sin40°
=4{cos(60) - cos100°} sin40°
=4×1/2sin40° - 4cos100°.sin40°
=2sin40° - 2{2sin40°cos100°}
=2sin40° -2{ sin140° -sin60° }
=2sin40° -2sin40° + 2sin60°
=√3 = RHS
khushnaved97:
yaar last mein -2sin40 kaha se aya wo to -sin140° hoga na
bolo
Sin 40°= sin ( 180° - 40°)
Sorry (180°-140°)
bhai complementary mein to 90-theta hota 180 kaha se layr
Associated angle
yaar associate mein esha to rule nhi tha (180-theta)
Bhai koi mera question bta do .... Range of sin x - cos x
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