Find the exact value of tan2(Beta) if sin(Beta) = 5/13 (Beta in Quadrant II)
sin x = 5/13
so tan x = -5/12
tan 2x = 2tan x/(1 - tan^2 x)
= 2(-5/12)/(1 - 25/144)
= -120/119
To find the exact value of tan2(Beta), we can use the double-angle formula for tangent:
tan2(Beta) = 2tan(Beta) / (1 - tan^2(Beta))
First, we need to find the value of tan(Beta).
Since sin(Beta) = 5/13 and Beta is in Quadrant II, we know that sin(Beta) is positive in Quadrant II.
Using the Pythagorean identity:
sin^2(Beta) + cos^2(Beta) = 1
We can solve for cos(Beta):
(5/13)^2 + cos^2(Beta) = 1
25/169 + cos^2(Beta) = 1
cos^2(Beta) = 1 - 25/169
cos^2(Beta) = 144/169
cos(Beta) = ±12/13
Since Beta is in Quadrant II, the cosine must be negative. Therefore, cos(Beta) = -12/13.
Now, we can find tan(Beta):
tan(Beta) = sin(Beta) / cos(Beta)
tan(Beta) = (5/13) / (-12/13)
tan(Beta) = -5/12
Finally, we can plug in the value of tan(Beta) into the double-angle formula:
tan2(Beta) = 2tan(Beta) / (1 - tan^2(Beta))
tan2(Beta) = 2(-5/12) / (1 - (-5/12)^2)
Simplifying further:
tan2(Beta) = -10/12 / (1 - 25/144)
tan2(Beta) = -5/6 / (119/144)
tan2(Beta) = (-5/6) * (144/119)
tan2(Beta) = -720/714
tan2(Beta) = -360/357
So the exact value of tan2(Beta) is -360/357.