Cos 2(alpha)=49/81and 2 alpha is in quad IV

Can someone help me with question. I honestly have no idea how to do this. Given cos(2alpha)=frac{49/81} and 2alpha is in quadrant IV, find exact values of the six trigonometric functions.
sin(alpha):
cos(alpha):
tan(alpha):
csc(alpha):
sec(alpha):
cot(alpha):

Let's find cos(α) first.

cos(2α) = 49/81
2cos^2(α) - 1 = 49/81 (because of the identity)
2cos^2(α) = 1 + 49/81 = 130/81
cos^2(α) = 65/81
cos(α) = ± √(65)/9

Now, 2α is in Q4, between 270º and 360º.
Therefore, α must be between 135º and 180º, which means it must be in Q2.
But cos is negative in Q2, meaning the cosine of an angle in Q2 is a negative number.
Thus, cos(α) = -√(65)/9.

Now find sin(α) from the identity : sin^2(α) + cos^2(α) = 1.
sin^2(α) = 1 - cos^2(α) = 1 - 65/81 = 16/81
sin(α) = ± 4/9

But sin is positive in Q2, so, sin(α) = 4/9.

Now that you have sin(α) = 4/9 and cos(α) = -√(65)/9, the rest should fall into place.

Note: there's a need to remember the following -
Sin is positive in Q1 and Q2. Cos is positive in Q1 and Q4.
If you just remember that, then you'll know which quadrants for each are negative.
You don't need to remember tan, because tan = sin/cos.