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

**M670** · Nov 24th 2012, 08:24 AM

Given

and

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

= .

= .

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

**skeeter** · Nov 24th 2012, 08:29 AM

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

**M670** · Nov 24th 2012, 08:32 AM

Originally Posted by skeeter

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

I realized that but this is the online question I need to answer from the university, my $\cos(2\alpha)$ should be a negative number if it is in the quad II

**Plato** · Nov 24th 2012, 08:33 AM

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

The $\cos$ is negative in quadrant II.

**M670** · Nov 24th 2012, 08:38 AM

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

**skeeter** · Nov 24th 2012, 08:40 AM

Originally Posted by M670

I realized that but this is the online question I need to answer from the university, my $\cos(2\alpha)$ 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.

**M670** · Nov 24th 2012, 08:42 AM

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** · Nov 24th 2012, 08:48 AM

your last step ...

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

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

**M670** · Nov 24th 2012, 08:53 AM

Originally Posted by skeeter

your last step ...

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

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

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

**skeeter** · Nov 24th 2012, 08:58 AM

no.

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

$\sin^2{\alpha} = \frac{9}{49}$

$\sin{\alpha} = \pm \frac{3}{7}$

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

**M670** · Nov 24th 2012, 09:34 AM

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

**skeeter** · Nov 24th 2012, 09:48 AM

if ...

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

then ...

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

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

**Plato** · Nov 24th 2012, 09:57 AM

Originally Posted by skeeter

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

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 $\alpha\in II$ . That would make all the information consistent and tell us $2\alpha\in IV$ .

**M670** · Nov 24th 2012, 10:10 AM

Well I just put inputted $\sin(\alpha)=\frac{3}{7}$ 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..

**M670** · Nov 24th 2012, 10:24 AM

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

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