If cos of theta = 3/5 and q terminates in the first quadrant, find the exact value of sin 2q. q=theta btw

Givens: cos(q) = 3/5

Sin(2q) = 2 sin(q)*cos(q)

Step One

Find sin(q)

Sin(q)= sqrt(1 - cos^2(q) )

Sin(q) = sqrt(1 - (3/5)^2 ) = sqrt(1 - (9/25) = sqrt( (16/25) ) = 4/5. Because the angle is in the first quadrant, sin(q) > 0.

Step 2

Find the value of sin(2q)

sin(2q) = 2 sin(q) * cos(q)

Sin(2q) = 2* 3/5 * 4/5

sin(2q) = 24/25.

Note: you should notice that sin(2q) is less than 1 which is the way a sine function should behave.

its c or d   take a lucky guess my dude