Find the exact value of the trigonometric function at the given real number. NS

Clifland 2021-10-22 Answered
Find the exact value of the trigonometric function at the given real number.
\(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\)

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Expert Answer

Malena
Answered 2021-10-23 Author has 17712 answers
We would like to find the exact value of \(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\), First, we can find the reference angle of \(\displaystyle{\frac{{{5}\pi}}{{{4}}}}\) where \(\displaystyle{\frac{{{5}\pi}}{{{4}}}}-{\frac{{\pi}}{{{4}}}}\), so the reference angle is \(\displaystyle{\frac{{\pi}}{{{4}}}}\).
Now the next step is to define the sign of \(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\). We know that \(\displaystyle{\frac{{{5}\pi}}{{{4}}}}\) is in quadrant 3 which the sine function is negative in this quadrant, so the value of \(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\) is negative and we can simplify it as follows:
\(\displaystyle\therefore{\sin{{\frac{{{5}\pi}}{{{4}}}}}}={\sin{{\left(\pi+{\frac{{\pi}}{{{4}}}}\right)}}}=-{\sin{{\frac{{\pi}}{{{4}}}}}}\)
Note that the first step was to find the reference angle and then was to define the sign of \(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\).
Now we can find the value of \(\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}\) by knowing the value of \(\displaystyle{\sin{{\frac{{\pi}}{{{4}}}}}}\) which equals \(\displaystyle{\frac{{\sqrt{{{2}}}}}{{{2}}}}\)
\(\displaystyle\therefore{\sin{{\frac{{{5}\pi}}{{{4}}}}}}={\sin{{\left(\pi+{\frac{{\pi}}{{{4}}}}\right)}}}=-{\sin{{\frac{{\pi}}{{{4}}}}}}=-{\frac{{\sqrt{{{2}}}}}{{{2}}}}\)
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