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4. Given a circle of radius 6 with the sector shaded as shown in the figure. (a) Find the angle \( \theta \) subtended at the center when arc length is \( s = 2\pi \). (b) Find the area of the sector formed by the arc \( s \). (c) Find the area of the triangle formed by the two radii \( r \). (d) Find the area of the shaded region. *** The figure shows a circle with radius \( r = 6 \). A sector is highlighted, with the angle \( \theta \) at the center and the arc length marked as \( s = 2\pi \). The two radii forming the sector meet the circle at the arc's endpoints. The shaded region is the area of interest. 3. Given \( f(x) = 4 \sin(2x - \frac{\pi}{2}) + 1 \) (a) Put it in standard form \( y = a \cos[k(x - b)] + v \). (b) Find the amplitude, period and crunch factor, the horizontal, and vertical shift. (c) Find the interval for one cycle, build a table of values and graph the function.

4. Given a circle of radius 6 with the sector shaded as shown in the figure. (a) Find the angle \( \theta \) subtended at the center when arc length is \( s = 2\pi \). (b) Find the area of the sector formed by the arc \( s \). (c) Find the area of the triangle formed by the two radii \( r \). (d) Find the area of the shaded region. *** The figure shows a circle with radius \( r = 6 \). A sector is highlighted, with the angle \( \theta \) at the center and the arc length marked as \( s = 2\pi \). The two radii forming the sector meet the circle at the arc's endpoints. The shaded region is the area of interest. 3. Given \( f(x) = 4 \sin(2x - \frac{\pi}{2}) + 1 \) (a) Put it in standard form \( y = a \cos[k(x - b)] + v \). (b) Find the amplitude, period and crunch factor, the horizontal, and vertical shift. (c) Find the interval for one cycle, build a table of values and graph the function.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section: Chapter Questions
Problem 77RE
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4. Given a circle of radius 6 with the sector shaded as shown in the figure. (a) Find the angle \( \theta \) subtended at the center when arc length is \( s = 2\pi \). (b) Find the area of the sector formed by the arc \( s \). (c) Find the area of the triangle formed by the two radii \( r \). (d) Find the area of the shaded region. *** The figure shows a circle with radius \( r = 6 \). A sector is highlighted, with the angle \( \theta \) at the center and the arc length marked as \( s = 2\pi \). The two radii forming the sector meet the circle at the arc's endpoints. The shaded region is the area of interest.
Transcribed Image Text:4. Given a circle of radius 6 with the sector shaded as shown in the figure. (a) Find the angle \( \theta \) subtended at the center when arc length is \( s = 2\pi \). (b) Find the area of the sector formed by the arc \( s \). (c) Find the area of the triangle formed by the two radii \( r \). (d) Find the area of the shaded region. *** The figure shows a circle with radius \( r = 6 \). A sector is highlighted, with the angle \( \theta \) at the center and the arc length marked as \( s = 2\pi \). The two radii forming the sector meet the circle at the arc's endpoints. The shaded region is the area of interest.
3. Given \( f(x) = 4 \sin(2x - \frac{\pi}{2}) + 1 \) (a) Put it in standard form \( y = a \cos[k(x - b)] + v \). (b) Find the amplitude, period and crunch factor, the horizontal, and vertical shift. (c) Find the interval for one cycle, build a table of values and graph the function.
Transcribed Image Text:3. Given \( f(x) = 4 \sin(2x - \frac{\pi}{2}) + 1 \) (a) Put it in standard form \( y = a \cos[k(x - b)] + v \). (b) Find the amplitude, period and crunch factor, the horizontal, and vertical shift. (c) Find the interval for one cycle, build a table of values and graph the function.
Expert Solution
Step 1

4).

As we know that the expression for the arc length of the circle is:

s=r×θ

Where r is the radius of the circle, θ is the angle subtended at the center.

(a).

Given:

r=6 ; s=2π

Now recall the above expression and substitute these values.

s=r×θ2π=6×θθ=2π6θ=π3

Therefore,

The angle subtended at the center: θ=π3

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