Find the value of:sin(-300°)cos^2 (120) + cos^2(-240 )sin^3(370)
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Answer:
=-sin300cos^2(180-60)+(-cos240)^2sin^3(360+10)
=-sin(360-60)(-1/2)^2+{(cos(180+60)}^2sin^3(10)
=sin60(1/4)+{-1/2}^2.sin^3(10)
What value of sin 10 to be used?
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