First, let's use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to find the value of sine.
Since the angle is in Quadrant III, both sine and cosine will be negative.
Given that cos(θ) = -16√, we can square both sides to find sin(θ):
sin^2(θ) + (-16√)^2 = 1
sin^2(θ) + 256 = 1
sin^2(θ) = 1 - 256
sin^2(θ) = -255
Since sin(θ) can't be negative, we know that the equation has no real solutions in this case. Therefore, there is no tangent value we can find.
Blake is told that an angle has a cosine of −16√
and is in Quadrant III. Use the Pythagorean identity sin2(θ)+cos2(θ)=1
, the trigonometric identity tanθ=sinθcosθ
, and the quadrant to find the tangent. (1 point)
1 answer
1 answer