Find cos 2A if tan A = 5/12 and <A lies in quadrant III.
If A is in quadrant 3, then 2 A is in quadrant 1 or 2
5^2 + 12^2 = 169 = 13^3
so sin A = -5/13
and cos A = -12/13
cos 2A = cos^2 A - sin^2 A
= (144 - 25)/169 = 119/169
To find cos 2A, we need to know the value of cos A. Since we are given tan A = 5/12, we can use the tangent identity to find cos A.
The tangent identity states that tan A = sin A / cos A.
Substituting the given value, we have:
5/12 = sin A / cos A
Next, we can solve this equation for sin A:
5/12 * cos A = sin A
Using the Pythagorean identity (sin^2 A + cos^2 A = 1), we can substitute sin A:
5/12 * cos A = √(1 - cos^2 A)
Simplifying this equation:
25/144 * cos^2 A = 1 - cos^2 A
25/144 * cos^2 A + cos^2 A = 1
(25/144 + 1) * cos^2 A = 1
169/144 * cos^2 A = 1
cos^2 A = 144/169
Since <A lies in quadrant III and cos A is negative in that quadrant, we take the negative square root:
cos A = -12/13
Now that we know the value of cos A, we can find cos 2A using the double-angle identity for cosine:
cos 2A = cos^2 A - sin^2 A
Substituting the values we have derived:
cos 2A = (-12/13)^2 - (5/12)^2
cos 2A = 144/169 - 25/144
cos 2A = (144/169) - (25/144)
cos 2A = (144*144 - 169*25)/(169 * 144)
cos 2A = 16711/27072
Therefore, cos 2A is equal to 16711/27072.