Given that sin(θ)=-1/squareroot2,how do you find the two possible angles for θ, given that −π<θ≤2π?

2 Answers
Jun 6, 2018

sin t = - 1/sqrt2
Trig table and unit circle give 2 solutions for t.
t = - pi/6, and t = pi - (-pi/6) = pi + pi/6 = (7pi)/6
Note.
t = - pi/6 is co-terminal to t = (11pi)/6
t = (7pi)/6 is co-terminal to t = - (5pi)/6
Inside the interval (-pi, 0) the answers are:
t = - (5pi)/6 , and t = - pi/6
Inside the interval (0, 2pi), the answers are:
t = (7pi)/6, and t = (11pi)/6

Jun 7, 2018

Here,

sintheta=-1/sqrt2 ,where,-pi < theta <=2pi

We have,

sintheta=-1/sqrt2 < 0 =>color(red)(III^(rd)Quadrant) orcolor(blue)( IV^(th)Quadrant)

Now,

-pi < theta <=2pi=>-pi < theta < 0 or 0 <= theta <=2pi

We have two options :

(i)-pi < theta < 0 ,then

sintheta=-1/sqrt2

=>theta=-pi+pi/4=-(3pi)/4tocolor(red)(III^(rd)Quadrant

or theta=0-pi/4=-pi/4tocolor(blue)(IV^(th)Quadrant)

(ii)0 <=theta <=2pi

sintheta=-1/sqrt2

=>theta=pi+pi/4=(5pi)/4tocolor(red)(III^(rd)Quadrant

or theta=2pi-pi/4=(7pi)/4tocolor(blue)(IV^(th)Quadrant)

Hence,

theta=-(3pi)/4,-pi/4, or (5pi)/4,(7pi)/4