1. ## Simple Trigonometry Help

Hey umm how do i find the degrees for sin thetha = - sqrt of 3 /2

The degree answer in my textbook says 240 degrees?

2. Originally Posted by MP3
Hey umm how do i find the degrees for sin thetha = - sqrt of 3 /2

The degree answer in my textbook says 240 degrees?
You should know that the reference angle is $\displaystyle \sin^{-1} \frac{\sqrt{3}}{2} = 60^0$. Use the reference angle to get the required angle that lies in the third quadrant (book's answer) and also fourth quadrant (omitted by book ....)

3. Thanks, i think i understand the concept of this but i have one more question.

What if the question was Cos theta = -1/ sqrt2 ?
The reference angle would be 60 degrees? and it would be at quadrant 2?
So 180-60= 120degrees BUT my textbook says 135degrees. What might be the proplem?

Oh yea for the previous question i asked you, why does sin thetha = - sqrt of 3 /2 belong to the third and fourth quadrant?

Originally Posted by mr fantastic
You should know that the reference angle is $\displaystyle \sin^{-1} \frac{\sqrt{3}}{2} = 60^0$. Use the reference angle to get the required angle that lies in the third quadrant (book's answer) and also fourth quadrant (omitted by book ....)

4. $\displaystyle cos \Theta = - \frac {1}{\sqrt{2}} = -\frac {\sqrt{2}}{\sqrt{2}\times\sqrt{2}} = -\frac {\sqrt{2}}{2} \Rightarrow \Theta = 180^0 \pm 45^0$

1)
$\displaystyle cos^{-1} (-\frac {\sqrt{2}}{2}) = 135^0$

2)
$\displaystyle cos^{-1} (-\frac {\sqrt{2}}{2}) = 225^0$

5. Originally Posted by MP3

The reference angle would be 60 degrees?
No
because
$\displaystyle cos 45= \frac{1}{\sqrt{2}}$

Since cos is negative it can either be in 2nd or 3rd quadrant
so

$\displaystyle cos(180-45) =\frac{-1}{\sqrt{2}}$
and
$\displaystyle cos(180+45) =\frac{-1}{\sqrt{2}}$

6. Originally Posted by MP3

Oh yea for the previous question i asked you, why does sin thetha = - sqrt of 3 /2 belong to the third and fourth quadrant?
Code:
               y
2nd Q       |     1st Q
|
|
-----------------------------------x
0 |
3rd Q       |     4th Q
|
Sin is positive in the 1st and 2nd Quadrant.
Cos is positive in the 1st and 4th Quadrant.
Tan is positive in the 1st and 3rd Quadrant
Cot is positive in the 1st and 3rd Quadrant

so if you have an expression $\displaystyle cos^{-1} (-\frac {\sqrt{2}}{2})$ you know that the solution is in the 2nd and 3rd Quadrant because its negative there.

$\displaystyle sin \Theta = -\frac{\sqrt{3}}{2}$

From the minus you know that the solution will be in the 3rd and 4th Quadrant.

$\displaystyle \Rightarrow \Theta_{1} = 0^0 - 60^0 = -60^0 = 300^0$

$\displaystyle \Theta_{2} = 180^0 + 60^0 = 240^0$