How do you this and help me to find it?

Question

If the given is secθ=3 in Quadrant IV, then what's the cosθ, tanθ, sinθ, and cos(2θ)? Thank you. I appreciated it.

Answer 1

cosθ = 1/secθ = 1/3
sinθ = - (2/3)√2
tanθ = - 2√2
cos(2θ) = 2cos^2θ - 1
cos(2θ) = 2/9 - 1 = - 7/9

Answer 2

sec(theta) = h/a, and cos(theta) = a/h. If the secant measure is 3, or 3/1, then the cosine measure reveals 1/3.

For the others, we will need to know the opposite side's length. Use pythagorean's theorem. c^2 = a^2 + b^2
c = 3, and a = 1. 9 = 1 + b^2, b = 2sqrt(2). But it is in the 4th quad. so it is negative.

Now we can obtain the tangent. tan(theta) = o/a = -2sqrt(2)/1, or -2sqrt(2).

Sin(theta) = o/h, which is simply -2sqrt(2)/3.

And, I am too tired to recall how to deal with cos(2theta), sorry.

Answer 3

well if sec is the rec. of cos, then cos0=1/3. and cos is sohCAHtoa... which is adjacent, hypotenuse. so therefore, the adjacent side is 1 and the hyp is 3... therefore sin0 is rad 8. tan is opp/adj, so opp is rad 8 and adj is 1... tan= 8. i dont remember how to do cos(2o) actually, so i hope that helped.