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Question: \[\mathbf{\cos}\mathbf{A}\mathbf{+}\mathbf{\cos}\mathbf{(}\mathbf{24}\mathbf{0}^{\mathbf{o}}\mathbf{...

cos⁡A+cos⁡(240o+A)+cos⁡(240o−A)=\mathbf{\cos}\mathbf{A}\mathbf{+}\mathbf{\cos}\mathbf{(}\mathbf{24}\mathbf{0}^{\mathbf{o}}\mathbf{+ A) +}\mathbf{\cos}\mathbf{(}\mathbf{24}\mathbf{0}^{\mathbf{o}}\mathbf{-}\mathbf{A) =}

A

cos⁡A\cos A

B

0

C

3sin⁡A\sqrt{3}\sin A

D

3cos⁡A\sqrt{3}\cos A

Answer

0

Explanation

Solution

cos⁡A+[2cos⁡240ocos⁡A]\cos A + \lbrack 2\cos 240^{o}\cos A\rbrack = cos⁡A+2(−cos⁡60o)cos⁡A\cos A + 2( - \cos 60^{o})\cos A

= cos⁡A[1−2(12)]=0\cos A\left\lbrack 1 - 2\left( \frac{1}{2} \right) \right\rbrack = 0.