Question

How do you evaluate \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)?\]

Answer 1

Hint:First we have to see that where is \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\] in the quadrant. So, we can go graphing \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\].Otherwise we can take the help of the co-terminals concept and get the points and then solve it.

Complete step by step answer:
According to the question, we will first check in which quadrant does \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\] lies. We will graph \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\]. If the angle is positive then, it will start in an anti-clockwise direction, else it will go in clockwise direction.As we can see that the angle is negative, so we will start from in clockwise direction. We always start from the x-axis. When we point \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\] in the graph, we get that \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\] comes under the 4th Quadrant.

Our point will be \[\dfrac{\pi }{4}\] and the coordinates are \[\left( {\dfrac{{\sqrt 2 }}{2}, - \dfrac{{\sqrt 2 }}{2}} \right)\]. If we notice then we can see that we have completed one revolution, so we need to always find out how far it goes after completing one round. So, we will write:
\[( - 2\pi + t)\]
Now, \[2\pi \] in terms of the denominator \[4\] is:
\[\sin \left( {\dfrac{{ - 8\pi }}{4} + t} \right)\]
Now, according to question, we find that:
\[ \Rightarrow \sin \left( {\dfrac{{ - 9\pi }}{4}} \right) = \sin \left( {\dfrac{{ - 8\pi }}{4} - \dfrac{\pi }{4}} \right)\]
When we see that we can write it as a period name, then we simply write it as:
\[ \Rightarrow \sin \left( {\dfrac{{ - 9\pi }}{4}} \right) = \sin \left( { - \dfrac{\pi }{4}} \right)\]
This makes it a lot simpler and easier.
When we see the coordinates, we can tell that:
\[ \therefore \sin \left( { - \dfrac{\pi }{4}} \right) = \left( { - \dfrac{{\sqrt 2 }}{2}} \right)\]
This is negative here because the y-axis is negative.

Note:To point \[\sin \left( {\dfrac{{ - 9\pi }}{4}} \right)\] in the graph, we have to see how many \[\pi \] over \[4\] are there. According to the question, we know that it is \[9\pi \] over \[4\]. So, we can divide the ‘x’ and ‘y’ axis into \[4\] parts, where each axis is getting \[4\] parts. After that we can calculate where \[9\pi \] over \[4\] lies, and then we can get our quadrant.