Find cos 0 and sin 0 , given tan 0 = 2/7 and cos 0<0

Answer 1

Given that tan 0 = 2/7, we can write:

tan 0 = 2/7
sin 0/cos 0 = 2/7
sin 0 = 2cos 0 / 7

We are also given that cos 0 < 0. Since cosine is negative, we know that the angle 0 lies in the second or third quadrant.

Let's use the Pythagorean identity sin^2θ + cos^2θ = 1 to solve for sin 0 and cos 0:

sin^2θ + cos^2θ = 1
(2cosθ/7)^2 + cos^2θ = 1
4cos^2θ/49 + cos^2θ = 1
(4cos^2θ + 49cos^2θ) / 49 = 1
53cos^2θ / 49 = 1
cos^2θ = 49/53
cosθ = -√(49/53)
cosθ = -7/√53

Since cosine is negative and the angle is in the second or third quadrant, and given that cosθ = -7/√53, we have:

cosθ = -7/√53
sinθ = 2(-7/√53) / 7
sinθ = -2/√53

Therefore, cos 0 = -7/√53 and sin 0 = -2/√53.