Question

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

Answer 1

`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`

Answer 2

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`