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Find two values of angle A between 2pi where sin A =
( - √(2 ) )/(2)

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Answer:

Explanation:

First of all, we know that


\sin (\pi)/(4)=\frac{\sqrt[]{2}}{2}

And knowing the unit circle, we know that the sine takes negative values in 3rd and 4th quadrants. Therefore, from the above value of the angle, If we go π radians counterclockwise, we encounter negative values of sine; hence,


\sin \lbrack(\pi)/(4)+\pi\rbrack=-\frac{\sqrt[]{2}}{2}
\rightarrow\sin (5\pi)/(4)=\frac{-\sqrt[]{2}}{2}

The second value of the angle that yields the above value for sine is found by adding π/2 radians to the angle above (we are now in the 4th quadrant)


\sin (5\pi)/(4)+(\pi)/(2)=-\frac{\sqrt[]{2}}{2}


\rightarrow\sin (7\pi)/(4)=-\frac{\sqrt[]{2}}{2}

Hence, the two values of angles between 0 and 2π are 5π/4 and 7π/4.

User ObiWanKenobi
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