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

Suman Cole 2021-10-15 Answered
I need a review or step by step explain how these two problems work.
\(\displaystyle{\cos{{\left({14}\pi-{0}\right)}}}\)
Determine the limit as x approaches the given x-coordinate and the continuity of the function at the that x-coordinate.
\(\displaystyle\lim_{{{x}\to{15}\frac{\pi}{{8}}}}{\cos{{\left({4}{x}-{\cos{{\left({4}{x}\right)}}}\right)}}}\)

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Expert Answer

wheezym
Answered 2021-10-16 Author has 13806 answers
To find: The value of \(\displaystyle{\cos{{\left({14}\pi-{0}\right)}}}\).
Concept used: The value of \(\displaystyle{\cos{{2}}}\pi{n}\) is 1 where the value of n is 0,1,2,3...
Calculation: Find the value of \(\displaystyle{\cos{{\left({14}\pi-{0}\right)}}}\) as,
\(\displaystyle{\cos{{\left({14}\pi-{0}\right)}}}={\cos{{\left({14}\pi\right)}}}\)
\(\displaystyle={\cos{{\left({7}{\left({2}\pi\right)}\right)}}}\)
\(\displaystyle={1}\)
Answer: The value of \(\displaystyle{\cos{{\left({14}\pi-{0}\right)}}}\) is 1.
To find: \(\displaystyle\lim_{{{x}\to{\frac{{{15}\pi}}{{{8}}}}}}{\cos{{\left({4}{x}-{\cos{{\left({4}{x}\right)}}}\right)}}}\)
The trigonometric identity is \(\displaystyle{\cos{{\left({x}\right)}}}={\sin{{\left({\frac{{\pi}}{{{2}}}}-{x}\right)}}}\) and \(\displaystyle{\sin{{\left(-{x}\right)}}}=-{\sin{{\left({x}\right)}}}\)
Calculation:
Solve by plugging \(\displaystyle{x}={\frac{{{15}\pi}}{{{8}}}}\) as,
\(\displaystyle\lim_{{{x}\to{\frac{{{15}\pi}}{{{8}}}}}}{\cos{{\left({4}{x}-{\cos{{\left({4}{x}\right)}}}\right)}}}={\cos{{\left({4}\cdot{\frac{{{15}\pi}}{{{8}}}}-{\cos{{\left({4}\cdot{\frac{{{15}\pi}}{{{8}}}}\right)}}}\right)}}}\)
\(\displaystyle={\cos{{\left({\frac{{{15}\pi}}{{{2}}}}-{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}\right)}}}\right)}}}\)
Find the value of \(\displaystyle{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}\right)}}}\) as,
\(\displaystyle{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}\right)}}}={\sin{{\left({\frac{{\pi}}{{{2}}}}-{\frac{{{15}\pi}}{{{2}}}}\right)}}}\)
\(\displaystyle={\sin{{\left(-{7}\pi\right)}}}\)
\(\displaystyle=-{\sin{{\left(-{7}\pi\right)}}}\)
\(\displaystyle={0}\)
Now apply the value \(\displaystyle{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}\right)}}}={0}\) in the expression \(\displaystyle{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}-{\cos{{\left({\frac{{{15}}}{{\pi}}}\right)}}}\right)}}}\)
\(\displaystyle{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}-{\cos{{\left({\frac{{{15}\pi}}{{{2}}}}\right)}}}\right)}}}={\cos{{\left({\frac{{{15}\pi}}{{{2}}}}-{0}\right)}}}\)
Answer: The value of \(\displaystyle\lim_{{{x}\to{\frac{{{15}\pi}}{{{8}}}}}}{\cos{{\left({4}{x}-{\cos{{\left({4}{x}\right)}}}\right)}}}\) is 0.
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