# 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. = . = . = . = . = . = .

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 $$\displaystyle \cos(2\alpha) > 0$$

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#### 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

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

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

#### M670

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

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?