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

A. 0
B. 1
C. 1/2
D. -3/5

Just noticed that you posted this new question after Steve showed you how to do it.

In the one Steve answered for you, it had asked for sin(2A)
This time they are asking for cos(2A), surely you can make that minor adjustment.
Did you even look at his solution?

To find cos(2A), we can use the double-angle identity for cosine:

cos(2A) = cos^2(A) - sin^2(A)

Since we are given that tan(A) = 2 and A is in Quadrant I, we can use the following identities:

tan(A) = sin(A) / cos(A)

From this, we can deduce that sin(A) = 2cos(A).

Substituting this value in the identity for cos(2A), we have:

cos(2A) = cos^2(A) - (2cos(A))^2

Simplifying this equation, we have:

cos(2A) = cos^2(A) - 4cos^2(A)

Factoring out cos^2(A), we get:

cos(2A) = cos^2(A)(1 - 4)

Now, we'll use the identity cos^2(A) + sin^2(A) = 1. Since A is in Quadrant I, both cos(A) and sin(A) are positive, so we have:

1 = cos^2(A) + (2cos(A))^2
1 = cos^2(A) + 4cos^2(A)

Combining like terms, we get:

1 = 5cos^2(A)
cos^2(A) = 1/5

Taking the square root of both sides, we have:

cos(A) = ± √(1/5)

Since A is in Quadrant I, cos(A) is positive. So, we have:

cos(A) = √(1/5)

Substituting this value back into the equation for cos(2A), we get:

cos(2A) = (√(1/5))^2 - 4(√(1/5))^2

Simplifying, we have:

cos(2A) = 1/5 - 4/5

cos(2A) = -3/5

Therefore, the answer is option D, -3/5.