cos(0)=2/2 ,and <0<2pi, evaluate sin(0) and tan(0). sin (0)=? tan(0)=?
PLEASE HELP!!!!

Answer:
Step-by-step explanation:
The angle is between 3pi/2 and 2pi so it is in the fourth quadrant where both sin and tan are negative.
Given cos(θ) = sqrt(2)/2
cos^2(θ) + sin^2(θ) = 1
sine^2(θ) = 1 - 2/2^2 = 1 - 1/2 = 1/2
sin(θ) = +/1 sqrt(2)/2
As we know sin(θ) is negative,
sin(θ) = -sqrt(2)/2; the second answer
tan(θ) = sin(θ)/cos(θ) = -1
Answer:
Step-by-step explanation:
using (cosθ)^2 + (sinθ)^2 = 1
2/4 + (sinθ)^2 = 1
sinθ = [tex]\sqrt{2}/2[/tex]
or -[tex]\sqrt{2} /2[/tex]
By C-A-S-T, sinθ is -ve
sinθ = -[tex]\sqrt{2}/2[/tex]
tanθ = sinθ/cosθ = -1