Find the Cos(2alpha) when its in the 2nd quadrant

Sep 2012
142
0
Montreal
Given
and
is in quadrant II, find exact values of the six trigonometric functions.

= .

= .

= .
= .

= .

= .

I know I need to set \(\displaystyle cos(2\alpha)=1-2\sin^2(\alpha)\) which is then \(\displaystyle \frac{31}{49}=1-2\sin^2(\alpha)\) which is then \(\displaystyle 2\sin^2(\alpha)=1-\frac{31}{49}\) now I think I get \(\displaystyle 2\sin^2(\alpha)= \frac{49}{49}-\frac{39}{49}=\frac{18}{49}\) from here I need some help ?\(\displaystyle 2\sin^2(\alpha)=\frac{18}{49}\)
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
something's wrong here ... if \(\displaystyle 2\alpha\) is in quad II , then \(\displaystyle \cos(2\alpha) < 0\). Your posted value for \(\displaystyle \cos(2\alpha) > 0\)
 
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Sep 2012
142
0
Montreal
something's wrong here ... if \(\displaystyle 2\alpha\) is in quad II , then \(\displaystyle \cos(2\alpha) < 0\). Your posted value for \(\displaystyle \cos(2\alpha) > 0\)
I realized that but this is the online question I need to answer from the university, my \(\displaystyle \cos(2\alpha)\) should be a negative number if it is in the quad II
 

Plato

MHF Helper
Aug 2006
22,492
8,653
Given
and
is in quadrant II, find exact values of the six trigonometric functions.

There is a serious flaw with this question.

If \(\displaystyle 2\alpha\in II\) then it is impossible that \(\displaystyle \cos(\2\alpha)=\frac{31}{49}\).

The \(\displaystyle \cos\) is negative in quadrant II.
 
Sep 2012
142
0
Montreal
Is there anyway to solve this problem as it's due for Monday? If say I tried to solve for \(\displaystyle \cos(2\alpha)=-\frac{31}{49}\)
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I realized that but this is the online question I need to answer from the university, my \(\displaystyle \cos(2\alpha)\) should be a negative number if it is in the quad II
... then you need to contact the responsible someone at your university and let them know the error.
 
Sep 2012
142
0
Montreal
See this is a bit of a problem I am at a university in Montreal that is using this program developed and maintained by the university of Rochester, so getting a response is very slow....
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
your last step ...

\(\displaystyle 2\sin^2{\alpha} = \frac{18}{49}\)

so ... \(\displaystyle \sin{\alpha} = \, ?\)
 
Sep 2012
142
0
Montreal
your last step ...

\(\displaystyle 2\sin^2{\alpha} = \frac{18}{49}\)

so ... \(\displaystyle \sin{\alpha} = \, ?\)
I believe it would be \(\displaystyle \sin{\alpha} = \frac{\frac{\sqrt18}{49}}{2}\)
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
no.

\(\displaystyle 2\sin^2{\alpha} = \frac{18}{49}\)

\(\displaystyle \sin^2{\alpha} = \frac{9}{49}\)

\(\displaystyle \sin{\alpha} = \pm \frac{3}{7}\)

now ... you have to determine which is the correct value (the plus or the minus value). how will you determine that?