Question

Given tan(A) = 2 and A is in Quadrant I, find sin(2A) and find cos(2A).

Answer 1

We can use following formula

$tan^{2}(A) = \frac{1- cos(2A)}{1+cos(2A)} \\ \\ tan(A) =2, so \\ \\4= \frac{1- cos(2A)}{1+cos(2A)} \\ \\1-cos(2A)=4+4cos(2A) \\ \\ 5cos(2A) = -3 \\ \\ cos (2A) = -3/5$

Now, we need to find sin(2A).

$sin^{2}(2A) + cos^{2}(2A) = 1 \\ \\ sin^{2}(2A) + (- \frac{3}{5})^{2} =1 \\ \\ sin^{2}(2A) =1 - \frac{9}{25} \\ \\ sin^{2}(2A) = \frac{16}{25} \\ \\ sin(2A) = \frac{4}{5} \ or \ - \frac{4}{5}$

Because, the angle A is in the 1st quadrant, and cos(2A) is negative, that means that the angle 2A is in the 2nd quadrant and sin(2A) is a positive number,so sin(2A) = 4/5.


sin(2A) = 4/5
cos(2A) = -3/5