Prove that cos 570 × sin510 - sin330× cos 390 = 0
Answers
Answered by
141
we can write the following as below
cos570 = cos(360+210)
= cos210 = cos(180+30) = -√3/2.............(1)
sin510 = sin(360+150)
= sin150 = sin(180-30) = 1/2...................(2)
sin330 = sin(360-30)
= sin(-30) = -1/2..........................................(3)
cos390= cos(360+30)
= cos30 =√3/2.........................................(4)
now from all this value
substituting values in the give problem
{(1) × (2)} - {(3) × (4)}
={(-√3/2) × (1/2)} - {(-1/2) × (√3/2)}
={-√3/4} - {-√3/4}
=0............................................PROVED
cos570 = cos(360+210)
= cos210 = cos(180+30) = -√3/2.............(1)
sin510 = sin(360+150)
= sin150 = sin(180-30) = 1/2...................(2)
sin330 = sin(360-30)
= sin(-30) = -1/2..........................................(3)
cos390= cos(360+30)
= cos30 =√3/2.........................................(4)
now from all this value
substituting values in the give problem
{(1) × (2)} - {(3) × (4)}
={(-√3/2) × (1/2)} - {(-1/2) × (√3/2)}
={-√3/4} - {-√3/4}
=0............................................PROVED
Answered by
76
cos570°×sin510°-cos390°×sin330°
=cos(540°+30°)×sin(540°-30°) - cos(360°+30°)×sin(360°-30°)
=cos30°×sin30°-cos30°×sin30°
=0
Hence proved
=cos(540°+30°)×sin(540°-30°) - cos(360°+30°)×sin(360°-30°)
=cos30°×sin30°-cos30°×sin30°
=0
Hence proved
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