Real World Applications of Sine and Cosine wave applications

Sine and Cosine can be used in everyday life in various ways. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Surveying, navagation, and astronomy all rely on sin and cos for the position of objects and other calculations.Music is composed of waves of different frequencies and amplitudes and these can be described using sin/cos. in fact anything involving sound waves will rely on sine and cosine. Ballistic trajectories rely on sin/cos. in physics, sin.cos is used often. Satellites and space flights relies on calculations and conversions to polar coordinates.Gps and cellphones rely on triangulation and formulas involving sin/cos. Signal transmissions involve waves described with sin/cos. Thermal analysis is used to model how things get hot e.g oven,spacecraft. this equation is usually solved using sums of sin and cosine. Also evaporation and [recipitation averages incorporate sin and cosine. Many other means of science/math in the real world incorparate sin/cos in some shape or form.

yahooanswers.com Real life applications of sine and cosine applications/functions

Week 2 Blog Prompt

What types of things can the graphs of sine and cosine be used to model in the real world? Give an example. **Be sure to cite if you use a source!**

adv math

For word problems they have 3 formulas that can be used to solve the equations
s=r x theta
k=1/2 r^2 x theta
k= 1/2 rs

these signs also have alternate meanings:
theta= apparent size
s=diameter of an object
r=distance between an object

ex:1

a sector of a circle has a radius of 8 cm and a central angle of 60 degrees. find its approximate arc length and area.
1st. list the formula
r=8cm theta=60 degrees
s=? k=? s=r x theta
2nd: plug in the formula
s=(8)(60)
s=140 degrees
3rd.plug in the formula to find k
k=1/2 x 8 x 140
k=560 cm

The thing that i understood most about chapter seven is the word problems. it has three formulas that can be used.

1. s=r x theta
2. k= 1/2 r^2 x theta
3. k= 1/2rs

some of the word problem hints is:
theta = apparent size
s= diameter of an object
r= distance between the objects

example 1:

A sector of a circle has a radius of 7cm and central angle of 50 degrees. find its approxiomate arc length and area.
r= 7cm theta=50 degrees
s=? k=?

then plug into a formula:
s=(7)(50)
s= 350 degrees

then plug that into a formula to find k.
k=1/2(7)(350)
k= 1225cm

This past week we basically reviewed everything from the week before. One thing we reviewed was finding an inverse of an angle. Both sin-1 and arcsin mean the same thing. They are both used to find inverse. They are two ways to find inverses: by plugging it into your calculator or working it yourself by drawing a triangle picture. When finding an inverse by using your calculator it is important that you have your calculator set to degrees.

example:
1) sin-1 0.9
plug into your calculator sin-1 0.9 and your answer should be 64.2

On Friday we started 8-1. To solve for theta you have to get the trig function by itself then take the inverse. First you find where the angle is based on the trig function and whether it is positive or negative. By having to leave class early Friday I missed some of the examples on how to do these types of problems.

Find a Reference Angle

7- 4 Finding a Reference Angle

Follow these steps to find a reference angle:
1. Locate which quadrant the angle is in.
2. Decide whether it is positive or negative.
3. Subtract 180 until the absolute value of theta is between 0 and 90 degrees.
4. If your answer is a trig chart angle plug it in. If it’s not you leave it alone or you plug it into your calculator.

Trig Chart Angles:
0 degrees = 0
30 degrees = pi/6
45 degrees = pi/4
60 degrees = pi/3
90 degrees = pi/2

Here are some examples you can try:

Find the reference angles:

Example 1:

sin(690°) = -sin(30°) = -π/6

690° is in the IV quadrant. They are asking for sin, and in the IV quadrant sin is negative. So we subtract 180° continuously from 690°, until we arrive at 30°. We then plug 30° into –sin and get-π/6, our answer.

Example 2:

cos(400°)=cos(40°)

400° is in quadrant I, which for cos is positive. We continuously subtract 180° from 400°, until we get an angle between 0° and 90°, which is 40°. Our final answer is cos(40°).

Area of a Sector

A circle sector is the part of a circle enclosed by two radii and an arc. When finding the area of a sector of a circle, you are actually finding a fractional part of the circle. The percentage is equated by the ratio of the central angle to the entire central angle, which is 360 degrees. To find the area of a sector in degrees, you would do the following steps:

1. First, identify the degree of the sector

2.Then, identify the radius of the circle

3. Multiply the radius (squared) times theta times pie.

EXAMPLE:

Find the area of a sector with a central angle of 60 degrees and a radius of 10. Express answer to the nearest tenth.

A = (theta)(pie)(radius)[squared]

(60)(pie)(10)[squared]

52.35987756

A = 52.4

Finding the area of a sector is fairly easy and useful. When we understand the area of a sector we can see the importance of the relationship of pie, theta, and the radius of a circle. Understanding circle sectors can lead to far more knowledge when learning about circles.