Answer:
Terminal point (0, -1); sin Ø = -1 ⇒ A
Step-by-step explanation:
In the unit circle, Ф is the angle between the terminal side and the positive part of the x-xis
- The terminal point on the positive part of the x-axis is (1, 0),which means Ф = 0° or 360° and cosФ = 1, sinФ = 0
- The terminal point on the positive part of the y-axis is (0, 1),which means Ф = 90° and cosФ = 0, sinФ = 1
- The terminal point on the negative part of the x-axis is (-1, 0),which means Ф = 180° and cosФ = -1, sinФ = 0
- The terminal point on the negative part of the y-axis is (0, -1),which means Ф = 270° and cosФ = 0, sinФ = -1
In a unit circle
∵ Ф = 270°
→ By using the 4th rule above
∴ The terminal point is (0, -1)
∴ sinФ = -1
∴ Terminal point (0, -1); sin Ø = -1
Answer:
The standard form of the equation of the circle is 
.
Step-by-step explanation:
A circle is the set of points in a plane that lie a fixed distance, called the radius, from any point, called the center.
The equation of a circle in standard form is
where <em>r</em> is the radius of the circle, and <em>h</em>, <em>k</em> are the coordinates of its center.
When the center of the circle coincides with the origin 
, so
We are also told that the circle contains the point (0, 1), so we will use that information to find the radius <em>r</em>.
Therefore, the standard form of the equation of the circle is .
Answer:
see below
Step-by-step explanation:
y = 8x − 9
y = 4x − 1
Since both equations are equal to y, we can set them equal to each other
8x-9 = 4x-1
Subtract 4x from each side
8x-4x-9 = 4x-4x-1
4x-9 = -1
Add 9 to each side
4x-9+9 = -1+9
4x = 8
Divide each side by 4
4x/4 = 8/4
x=2
Now find y
y = 4x-1
y = 4(2)-1
y = 8-1
y = 7
(2,7)
The point of intersection when the 2 lines are graphed is (2,7)
Answer:
2nd statement is false
Step-by-step explanation:
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
cot(x)sec⁴(x) cot(x)sec⁴(x)
0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
0 = cos⁴(x)(1 + tan²(x))²
0 = cos⁴(x) or 0 = (1 + tan²(x))²
⁴√0 = ⁴√cos⁴(x) or √0 = (√1 + tan²(x))²
0 = cos(x) or 0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x or √-1 = √tan²(x)
i = tan(x)
(No Solution)
2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
sin²(x) - cos²(x) = sin²(x) - cos²(x)
+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
0 = 0
3. 1 + sec²(x)sin²(x) = sec²(x)
sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
x = 0
4. -tan²(x) + sec²(x) = 1
-1 -1
tan²(x) - sec²(x) = -1
tan²(x) = -1 + sec²
√tan²(x) = √(-1 + sec²(x))
tan(x) = √(-1 + sec²(x))
tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
x = 0
