Hey umm how do i find the degrees for sin thetha =  sqrt of 3 /2
The degree answer in my textbook says 240 degrees?
Printable View
Hey umm how do i find the degrees for sin thetha =  sqrt of 3 /2
The degree answer in my textbook says 240 degrees?
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 18060= 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?
$\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
$
Sin is positive in the 1st and 2nd Quadrant.Code:y
2nd Q  1st Q


x
0 
3rd Q  4th Q

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