Simple Trigonometry Help

• January 24th 2009, 09:18 PM
MP3
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?
• January 24th 2009, 09:59 PM
mr fantastic
Quote:

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 $\sin^{-1} \frac{\sqrt{3}}{2} = 60^0$. 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 ....)
• January 25th 2009, 08:47 AM
MP3
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?

Quote:

Originally Posted by mr fantastic
You should know that the reference angle is $\sin^{-1} \frac{\sqrt{3}}{2} = 60^0$. 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 ....)

• January 25th 2009, 09:01 AM
metlx
$
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)
$
cos^{-1} (-\frac {\sqrt{2}}{2}) = 135^0
$

2)
$
cos^{-1} (-\frac {\sqrt{2}}{2}) = 225^0
$
• January 25th 2009, 09:04 AM
(Hi)
Quote:

Originally Posted by MP3

The reference angle would be 60 degrees?

No
because
$
cos 45= \frac{1}{\sqrt{2}}
$

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

$cos(180-45) =\frac{-1}{\sqrt{2}}$
and
$cos(180+45) =\frac{-1}{\sqrt{2}}$
• January 25th 2009, 11:54 AM
metlx
Quote:

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

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

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

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

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