find the values of the requested trig functions using the given value and quadrant in which the point corresponding to the angle lies.
quadrant II; sinθ=4/5
find cosθ and tan θ
in QII, x is negative, so cosθ = -3/5
The rest is easy, right?
i don't understand where you got the -3/5 from
better review your trig functions
sinθ = x/r
cosθ = y/r
r^2 = x^2+y^2
You have a 3-4-5 triangle.
Draw the dang thing!
x = -3, so cosθ = -3/5
To find the values of cosine (cosθ) and tangent (tanθ) given the value of sine (sinθ) and the quadrant in which the point corresponding to the angle lies, you can use the Pythagorean identity and the definition of tangent.
In Quadrant II, sine is positive and cosine is negative.
1. Start by finding the value of cosine (cosθ):
Since sinθ = 4/5, we know that the opposite side of the angle is 4 and the hypotenuse is 5 (according to the definition of sine as the ratio of the opposite side to the hypotenuse).
Using the Pythagorean identity: cosθ = √(1 - sin²θ)
Substitute the value of sinθ: cosθ = √(1 - (4/5)²)
cosθ = √(1 - 16/25)
cosθ = √(9/25)
cosθ = 3/5 (since cosθ is negative in Quadrant II)
2. Next, find the value of tangent (tanθ):
Tanθ is defined as the ratio of the opposite side (4) to the adjacent side (cosθ).
Therefore, tanθ = sinθ / cosθ
tanθ = (4/5) / (3/5)
tanθ = 4/3
In Quadrant II, cosθ = -3/5 and tanθ = 4/3.