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Cos³340+cos³390+cos³450please give correct answer |
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Answer» 40 o −3sin50 o ⇒cos 3 40 o −3sin(90−40)⇒cos 3 40−3cos40Let cos40=cosxby complementing trigonometric functionWe can write cos(n 2π ±θ)=cosθNow if cosx=cos40then cos3x=cos120Now ⇒cos3x=cos(2x+x)=cos2xcosx−sin2xsinx⇒cos120=(cos 2 x−sin 2 x)cosx−2sin 2 xcosx⇒ 2−1 =cos 3 x−(1−cos 2 x)cosx−2sin 2 xcosx⇒ 2−1 =cos 3 x−cosx+cos 3 x−2(1−cos 2 x)cosx⇒ 2−1 =2COS 3 x−cosx−2cosx+2cos 3 x⇒8cos 3 x−6cosx+1=0Solving this we getcosx=−0.94,0.174.0.766as cosx cannot be NEGATIVE between [0, 2π ] and always DECREASING i.e cos45 |
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