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Which of the following is not equivalent to 
cos ((pi)/(5)) ?

cos ((11 pi)/(5))

cos(-(9pi)/(5))

cos ((14 pi)/(5))

cos ((9pi)/(5))

Which of the following is not equivalent to cosπ5 \cos \frac{\pi}{5} ?\newlinecos11π5 \cos \frac{11 \pi}{5} \newlinecos(9π5) \cos \left(-\frac{9 \pi}{5}\right) \newlinecos14π5 \cos \frac{14 \pi}{5} \newlinecos9π5 \cos \frac{9 \pi}{5}

Full solution

Q. Which of the following is not equivalent to cosπ5 \cos \frac{\pi}{5} ?\newlinecos11π5 \cos \frac{11 \pi}{5} \newlinecos(9π5) \cos \left(-\frac{9 \pi}{5}\right) \newlinecos14π5 \cos \frac{14 \pi}{5} \newlinecos9π5 \cos \frac{9 \pi}{5}
  1. Properties of Cosine Function: Understand the properties of the cosine function. The cosine function has a period of 2π2\pi, which means that cos(θ)=cos(θ+2πk)\cos(\theta) = \cos(\theta + 2\pi k) for any integer kk. Also, cosine is an even function, so cos(θ)=cos(θ)\cos(\theta) = \cos(-\theta).
  2. Evaluate cos(11π5)\cos \left(\frac{11 \pi}{5}\right): Evaluate cos(11π5)\cos \left(\frac{11 \pi}{5}\right). Using the periodic property of cosine, we can subtract 2π2\pi (which is the same as 10π5\frac{10\pi}{5}) from 11π5\frac{11\pi}{5} to find an equivalent angle within the first period. cos(11π5)=cos(11π510π5)=cos(1π5)\cos \left(\frac{11 \pi}{5}\right) = \cos \left(\frac{11 \pi}{5} - \frac{10 \pi}{5}\right) = \cos \left(\frac{1 \pi}{5}\right) Since 1π5\frac{1\pi}{5} is equivalent to π5\frac{\pi}{5}, this expression is equivalent to cos(π5)\cos \left(\frac{\pi}{5}\right).
  3. Evaluate cos(9π5)\cos \left(-\frac{9\pi}{5}\right): Evaluate cos(9π5)\cos \left(-\frac{9\pi}{5}\right). Using the even property of cosine, we know that cos(θ)=cos(θ)\cos(\theta) = \cos(-\theta). cos(9π5)=cos(9π5)\cos \left(-\frac{9\pi}{5}\right) = \cos \left(\frac{9\pi}{5}\right) Now, we can add 2π2\pi (which is the same as 10π5\frac{10\pi}{5}) to 9π5\frac{9\pi}{5} to find an equivalent angle within the first period. cos(9π5)=cos(9π5+10π5)=cos(19π5)\cos \left(\frac{9\pi}{5}\right) = \cos \left(\frac{9\pi}{5} + \frac{10\pi}{5}\right) = \cos \left(\frac{19\pi}{5}\right) We can subtract 2π2\pi again to find an equivalent angle. cos(19π5)=cos(19π510π5)=cos(9π5)\cos \left(\frac{19\pi}{5}\right) = \cos \left(\frac{19\pi}{5} - \frac{10\pi}{5}\right) = \cos \left(\frac{9\pi}{5}\right) Since we have already established that cos(9π5)=cos(9π5)\cos \left(-\frac{9\pi}{5}\right) = \cos \left(\frac{9\pi}{5}\right), this expression is equivalent to cos(9π5)\cos \left(-\frac{9\pi}{5}\right)11.
  4. Evaluate cos(14π5)\cos \left(\frac{14 \pi}{5}\right): Evaluate cos(14π5)\cos \left(\frac{14 \pi}{5}\right). Using the periodic property of cosine, we can subtract 2π2\pi (which is the same as 10π5\frac{10\pi}{5}) from 14π5\frac{14\pi}{5} to find an equivalent angle within the first period. cos(14π5)=cos(14π510π5)=cos(4π5)\cos \left(\frac{14 \pi}{5}\right) = \cos \left(\frac{14 \pi}{5} - \frac{10 \pi}{5}\right) = \cos \left(\frac{4 \pi}{5}\right) Since 4π5\frac{4\pi}{5} is not equivalent to π5\frac{\pi}{5}, this expression is not equivalent to cos(π5)\cos \left(\frac{\pi}{5}\right).
  5. Evaluate cos(9π5)\cos \left(\frac{9\pi}{5}\right): Evaluate cos(9π5)\cos \left(\frac{9\pi}{5}\right). Using the periodic property of cosine, we can subtract 2π2\pi (which is the same as 10π5\frac{10\pi}{5}) from 9π5\frac{9\pi}{5} to find an equivalent angle within the first period. cos(9π5)=cos(9π510π5)=cos(1π5)\cos \left(\frac{9\pi}{5}\right) = \cos \left(\frac{9\pi}{5} - \frac{10\pi}{5}\right) = \cos \left(-\frac{1\pi}{5}\right) Using the even property of cosine, we know that cos(θ)=cos(θ)\cos(\theta) = \cos(-\theta). cos(1π5)=cos(1π5)\cos \left(-\frac{1\pi}{5}\right) = \cos \left(\frac{1\pi}{5}\right) Since 1π5\frac{1\pi}{5} is equivalent to π5\frac{\pi}{5}, this expression is equivalent to cos(9π5)\cos \left(\frac{9\pi}{5}\right)00.

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