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Question: Evaluate \[\cos \left( { - \dfrac{{8\pi }}{3}} \right)\] ?...

Evaluate cos⁡(−8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right) ?

Explanation

Solution

Use the trigonometric properties to find the value of cos⁡(−8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right).
Since cos⁡(−x)=cos⁡x\cos ( - x) = \cos x so evaluate cos⁡(−8π3)=cos⁡(8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right) = \cos \left( {\dfrac{{8\pi }}{3}} \right).
Write cos⁡(8π3)=cos⁡(2π+2π3)\cos \left( {\dfrac{{8\pi }}{3}} \right) = \cos \left( {2\pi + \dfrac{{2\pi }}{3}} \right) .
Then apply the trigonometric formula; cos⁡(2π+x)=cos⁡x\cos (2\pi + x) = \cos x.

Complete step by step answer:
We have to evaluate cos⁡(−8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right). The value of the negative angle of the cosine function is the same as the positive angle of the cosine function.
cos⁡(−x)=cos⁡x\cos ( - x) = \cos x
⇒cos⁡(−8π3)=cos⁡(8π3)\Rightarrow \cos \left( { - \dfrac{{8\pi }}{3}} \right) = \cos \left( {\dfrac{{8\pi }}{3}} \right)
Write the angle 8π3=2π+2π3\dfrac{{8\pi }}{3} = 2\pi + \dfrac{{2\pi }}{3}.
⇒cos⁡(8π3)=cos⁡(2π+2π3)\Rightarrow \cos \left( {\dfrac{{8\pi }}{3}} \right) = \cos \left( {2\pi + \dfrac{{2\pi }}{3}} \right)
According to the trigonometric formula; cos⁡(2π+x)=cos⁡x\cos (2\pi + x) = \cos x,
⇒cos⁡(2π+2π3)=cos⁡(2π3)\Rightarrow \cos \left( {2\pi + \dfrac{{2\pi }}{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)
We can split the angle 2π3\dfrac{{2\pi }}{3} as π−π3\pi - \dfrac{\pi }{3} .
⇒cos⁡(2π3)=cos⁡(π−π3)\Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)
Since cos⁡(π−θ)=−cos⁡θ\cos (\pi - \theta ) = - \cos \theta we get,
⇒cos⁡(π−π3)=−cos⁡(π3)\Rightarrow \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \left( {\dfrac{\pi }{3}} \right)
And,
⇒−cos⁡(π3)=−12\Rightarrow - \cos \left( {\dfrac{\pi }{3}} \right) = - \dfrac{1}{2}

Note: Another Method:
We have to evaluate cos⁡(−8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right).
Write the angle −8π3=−2π−2π3 - \dfrac{{8\pi }}{3} = - 2\pi - \dfrac{{2\pi }}{3}.
⇒cos⁡(8π3)=cos⁡(−2π−2π3)\Rightarrow \cos \left( {\dfrac{{8\pi }}{3}} \right) = \cos \left( { - 2\pi - \dfrac{{2\pi }}{3}} \right)
According to the trigonometric formula; cos⁡(−2π−x)=cos⁡(−x)\cos ( - 2\pi - x) = \cos ( - x),
⇒cos⁡(−2π−2π3)=cos⁡(−2π3)\Rightarrow \cos \left( { - 2\pi - \dfrac{{2\pi }}{3}} \right) = \cos \left( { - \dfrac{{2\pi }}{3}} \right)
Apply the property, cos⁡(−x)=cos⁡x\cos ( - x) = \cos x,
⇒cos⁡(−2π3)=cos⁡(2π3)\Rightarrow \cos \left( { - \dfrac{{2\pi }}{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)
⇒cos⁡(2π3)=−12\Rightarrow \cos \left( {\dfrac{{2\pi }}{3}} \right) = - \dfrac{1}{2}
The value of cos⁡(−8π3)\cos \left( { - \dfrac{{8\pi }}{3}} \right) is −12 - \dfrac{1}{2}.