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

Find the exact value of the trigonometric function at the given real number.
$$\displaystyle{\sin{{\frac{{{5}\pi}}{{{4}}}}}}$$

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Malena
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 $$\displaystyle{\frac{{\sqrt{{{2}}}}}{{{2}}}}$$
$$\displaystyle\therefore{\sin{{\frac{{{5}\pi}}{{{4}}}}}}={\sin{{\left(\pi+{\frac{{\pi}}{{{4}}}}\right)}}}=-{\sin{{\frac{{\pi}}{{{4}}}}}}=-{\frac{{\sqrt{{{2}}}}}{{{2}}}}$$