I need a review or step by step explain how these two problems work. cos⁡(14π−0) Determine the limit as x approaches the given x-coordinate and the continuity of the function at the that x-coordinate. limx→15π8cos⁡(4x−cos⁡(4x))

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I need a review or step by step explain how these two problems work.  cos⁡(14π−0)  Determine the limit as x approaches the given x-coordinate and the continuity of the function at the that x-coordinate.  limx→15π8cos⁡(4x−cos⁡(4x))

wheezym · Accepted Answer

To find: The value of cos⁡(14π−0).  Concept used: The value of cos⁡2πn is 1 where the value of n is 0,1,2,3...  Calculation: Find the value of cos⁡(14π−0) as,  cos⁡(14π−0)=cos⁡(14π)  =cos⁡(7(2π))  =1  Answer: The value of cos⁡(14π−0) is 1.  To find: limx→15π8cos⁡(4x−cos⁡(4x))  The trigonometric identity is cos⁡(x)=sin⁡(π2−x) and sin⁡(−x)=−sin⁡(x)  Calculation:  Solve by plugging x=15π8 as,  limx→15π8cos⁡(4x−cos⁡(4x))=cos⁡(4⋅15π8−cos⁡(4⋅15π8))  =cos⁡(15π2−cos⁡(15π2))  Find the value of cos⁡(15π2) as,  cos⁡(15π2)=sin⁡(π2−15π2)  =sin⁡(−7π)  =−sin⁡(−7π)  =0  Now apply the value cos⁡(15π2)=0 in the expression cos⁡(15π2−cos⁡(15π))  cos⁡(15π2−cos⁡(15π2))=cos⁡(15π2−0)  Answer: The value of limx→15π8cos⁡(4x−cos⁡(4x)) is 0.

I need a review or step by step explain how these two problems work. \cos

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