Find the value of tangent of angle A, given cos2A = 4/5, and A terminates in quadrant 2
I am not going to do them all for you. Let's see you try.
I have the same equation, so I will try. The "cos2A" is throwing me off. Since it is in QII it will be <90 and >180. Since it is a COS the value of theta should be x/r?
x in QII is negative so x=(-4)
r = 5
y = 3
Almost hate to go further if I am off
since TAN(theta)=y/x
TAN(theta)= 3/(-4)
here is where the "cos2A" throws me and I cant find any examples. Is it as simple as 2(3/5)?
Appreciate any clarity.
To find the value of tangent of angle A, we first need to determine the value of sin(A) and cos(A).
Given that A terminates in quadrant 2 and cos^2(A) = 4/5, we can find cos(A) by taking the square root of 4/5.
cos(A) = ±√(4/5)
Since A terminates in quadrant 2, cos(A) is negative. Therefore, cos(A) = -√(4/5).
Next, we can use the Pythagorean identity to find sin(A):
sin^2(A) + cos^2(A) = 1
By plugging in cos(A) = -√(4/5):
sin^2(A) + (-√(4/5))^2 = 1
sin^2(A) + (4/5) = 1
sin^2(A) = 1 - (4/5)
sin^2(A) = (5/5) - (4/5)
sin^2(A) = 1/5
Taking the square root of both sides:
sin(A) = ±√(1/5)
Since A terminates in quadrant 2, sin(A) is positive. Therefore, sin(A) = √(1/5).
Finally, we can use the tangent definition to find the value of tangent(A):
tan(A) = sin(A) / cos(A)
tan(A) = √(1/5) / -√(4/5)
tan(A) = -√(1/5) / √(4/5)
To simplify, we can rationalize the denominator:
tan(A) = -√(1/5) * √(5/4) / ( √(4/5) * √(5/4) )
tan(A) = -√(5/20) / √(20/20)
tan(A) = -√5 / √20
Finally, by simplifying the expression, we get the value of tangent(A):
tan(A) = -√(5/4)
Therefore, the value of tangent of angle A is -√(5/4).