Question

Evaluate the trigonometric function using its period as an aid. sin(-19pi/3)

Answer 1

If you add 2π or subtract 2π from an angle, you do not change its position along the unit circle, since 2π is just a full revolution.
Therefore sin(x+2π) = sin(x) = sin(x−2π)

Also, sin(x) is an odd function, so sin(−x) = −sin(x)
and cos(x) is an even function, so cos(−x) = cos(x)

cos(3π) = cos(3π−2π) = cos(π) = −1

sin(9π/4) = sin(9π/4−2π) = sin(π/4) = √2/2

cos(19π/6) = cos(19π/6−2π) = cos(7π/6) = −√3/2

sin(−13π/6) = sin(−13π/6+2π) = sin(−π/6) = −sin(π/6) = −1/2

cos(−8π/3) = cos(−8π/3+2π) = cos(−2π/3) = cos(2π/3) = −1/2

Answer 2

√2/2

Answer 3

s o 1/sqrt2..

Answer 4

medal pls

Answer 5

it was wrong )-: and you just copy and pasted lol