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

M670

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

skeeter

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

1 person

M670

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
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.

M670

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
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.

M670

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
your last step ...

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

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

M670

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
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?

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