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

Given and is in quadrant II, find exact values of the six trigonometric functions.

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

something's wrong here ... if $\displaystyle 2\alpha$ is in quad II , then $\displaystyle \cos(2\alpha) < 0$. Your posted value for $\displaystyle \cos(2\alpha) > 0$

Originally Posted by skeeter

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

Originally Posted by M670

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.

Is there anyway to solve this problem as it's due for Monday? If say I tried to solve for $\displaystyle \cos(2\alpha)=-\frac{31}{49}$

Originally Posted by M670

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.

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....

your last step ...

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

so ... $\displaystyle \sin{\alpha} = \, ?$

Originally Posted by skeeter

your last step ...

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

so ... $\displaystyle \sin{\alpha} = \, ?$

I believe it would be $\displaystyle \sin{\alpha} = \frac{\frac{\sqrt18}{49}}{2}$

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?

That would be determined by the fact the question is asking me to look in the 2nd quad and sin of an angle in QII is positive

if ...

$\displaystyle \frac{\pi}{2} < 2\alpha < \pi$

then ...

$\displaystyle \frac{\pi}{4} < \alpha < \frac{\pi}{2}$

... therefore, $\displaystyle \alpha$ is an angle in quad I

Originally Posted by skeeter

something's wrong here ... if $\displaystyle 2\alpha$ is in quad II , then $\displaystyle \cos(2\alpha) < 0$. Your posted value for $\displaystyle \cos(2\alpha) > 0$

Don't you think it is pointless to continue to speculate on what the correct question should have been?

I suspect that whoever wrote the question meant $\displaystyle \alpha\in II$. That would make all the information consistent and tell us $\displaystyle 2\alpha\in IV$.

Well I just put inputted $\displaystyle \sin(\alpha)=\frac{3}{7}$ and the answer is correct I received 17% out of 100 because I didn't answer the other 5 functions but I still have 1 more attempt to add in the rest of the answer correct..

I believe $\displaystyle \cos(\alpha)=\frac{x}{7}$ $\displaystyle X=\sqrt40$ so $\displaystyle \cos(\alpha)=\frac{\sqrt40}{7}$ Does this make sense?