vectors and angles

Here are some two dimensional vectors. Find the difference between the vectors, both U-V and V-U, the dot products U·U, V·V, U·V, length of the vectors and the cosine of the angle between them, which is given by the formula in the picture above.

Example

U = <4 3> and V = <0 5>

U·U =  25

V·V =  25

U·V = 15

||U|| = sqrt(25) = 5

||V|| = sqrt(25) = 5

costheta = 15/25 = 3/5

theta ~= 53.1301°

Practice.

U = <4 3> and V = <3 -4>

U·U = ______

V·V =  ______

U·V =______

||U|| = ______

||V|| = ________

costheta = ______

theta = _______

U-V = _______

||U-V|| = _________

cosine of angle between U and U-V = ________


angle to nearest thousandth of a degree = _______


V-U = _______

||V-U|| = _________

cosine of angle between V and V-U = ________


angle to nearest thousandth of a degree = _______



Answers in the comments.

Practice problems similar to take home

a) Rewrite cos^4(alpha) - sin^4(alpha) in a simpler form with lower powers of the trig functions

b) let cosgamma = 5/7 find
singamma
tangamma
cos(½gamma)
sin(½gamma)
tan(½gamma)
cos(2gamma)
sin(2gamma)
tan(2gamma)

c) The sides of a triangle are A = 5, B = 5 and C = 8.

Find the area.
Find the approximate value of alpha in degrees, rounded to four places after the decimal.
Find the approximate value of beta in degrees, rounded to four places after the decimal.
Find the approximate value of gamma in degrees, rounded to four places after the decimal.
Find the height if A is the base.
Find the height if B is the base.
Find the height if C is the base.

d) A = 6, B = 7, beta = 90° Find sinalpha and alpha written in degrees rounded four places after the decimal.

Answers in the comments.

Practice problems for Homework 10

Find the high point and low point surrounding the median point closest to (0, 0) for the following trig functions.

f(x) = 6sinx - 8cosx

g(x) = 2sinx - 4cosx

Here are three points on the unit circle in the complex plane.  Multiplying two points of the form costheta + isintheta and cosiota+isiniota gives us the values for the angle theta + iota.

a = 1/5 + i2sqrt(6)/5
b = 2/5 + isqrt(21)/5

ab =

a² =
a³ =

Answers in the comments. 

Examples for Homework 9

Half angle formulas:
cos½alpha = +/-sqrt((1+cosalpha)/2)

sin½alpha = +/-sqrt((1+cosalpha)/2)

The angle 315° is the same as (360-45)° or -45°, which means

sin315° = -sqrt(2)/2
cos315° = sqrt(2)/2

Half of 315 is 157.5°, which is in the second quadrant. Sine will be positive and cosine negative.

cos157.5° = -sqrt((1+sqrt(2)/2)/2

With a little algebraic manipulation, this becomes
-sqrt((2+sqrt(2))/4) and to have no radicals in the denominator the final answer is
cos157.5° = -sqrt(2+sqrt(2))/2

sin157.5° = sqrt((1-sqrt(2)/2)/2

With a little algebraic manipulation, this becomes
sqrt((2-sqrt(2))/4) and to have no radicals in the denominator the final answer is
sin157.5° = sqrt(2-sqrt(2))/2

157.5° in radians is (157.5/180)pi = (315/360)pi = (7/8)pi

===================

A start for cos3alpha

cos(alpha + 2alpha) = cosalpha×cos2alpha - sinalpha×sin2alpha
= cosalpha(cos²alpha - sin²alpha) - sinalpha(2cosalphasinalpha)

Continue the algebraic simplification.

Practice:Two consecutive extremes to f(x) = Asin(bx+c)+D f(x) = Asin(bx+c)+D to a middle point and the nearest maximum and minimum points

Two consecutive extremes

a) (6, 3) and (10, 4)

b) (-2, 1) and (4, -1)

c) (3, 5) and (5, 3)

Find the midpoint and the nearest max and min for the following functions

f(x) = 3sin([pi/2]x + 4) - 5


g(x) = -sin(x + pi/4) 


r(x) = -4sin(4x + 4) +4

answers in the comments.

The basics of graphing f(x) = Asin(bx+c) + D

  -->

Here is f(x) = sinx from x = -6 to 6.

Height (amplitude) = 1 (from 1 to -1)

Period = 2pi

f(0) = 0

f(pi/2) =1

f(-pi/2) = -1

Here is f(x) = 2sinx from x = -6 to 6.

Height (amplitude) = 2 (from 2 to -2)

Period = 2pi

f(0) = 0

f(pi/2) =2

f(-pi/2) = -2

The thing that is changed is the height or amplitude. f(x) = Asinx oscillates from A to -A. Choosing a negative A makes the graph start at 0 and move downward instead of upward.

  -->

Here is f(x) = sin2x from x = -6 to 6.

Height (amplitude) = 1 (from 1 to -1)

Period = pi

f(0) = 0

f(pi/4) =1

f(-pi/4) = -1

The thing that is changed is the period. f(x) = sinbx has period 2pi/b.

To repeat faster than 2pi, choose |b| > 1. For slower repeats |b| < 1. Choosing a negative b makes the graph start at 0 and move downward instead of upward.

Here is f(x) = sin(x + pi/4) from x = -6 to 6.

Height (amplitude) = 1 (from 1 to -1)

Period = 2pi

f(-pi/4) = 0

f(pi/4) =1

f(-3pi/4) = -1
What changes here are the x positions of the midpoint and the maximum and minimum values. f(x) = sin(x + c) reaches 0 at x = -c, the high point at x = pi/2 - c and the low point at x = -pi/2 - c.

Here is f(x) = sinx + pi/4 from x = -6 to 6.

Height (amplitude) = 1 (from 1+pi/4 to -1+pi/4)

Period = 2pi

f(0) =pi/4

f(pi/2) =1+pi/4

f(-pi/2) = -1+pi/4

What changes here are the y positions of the midpoint and the maximum and minimum values. Instead of oscillating between 1 and -1, f(x) = sinx + D goes back and forth between D+1 to D-1.


-->

To review.

f(x) = Asin(bx + c) + D

The constant A changes the height or amplitude of the graph, rising to |A| above the middle y value, which is the constant D, and falling to -|A| below the middle y value. Negative values of A cause the graph to go downward to the right of bx + c = 0 instead of upward.

The constant b changes the distance in x between the consecutive highest points, also known as the period of the function.  The standard period is 2pi, and if |b| ≠ 1, the period changes to 2pi/|b|. The constant c changes where on the x-axis the sine function reaches its middle value. This is called phase shift. The point of the middle value moves to bx + c  = 0, so positive values cause the graph to shift left and negative values cause the graph to shift right.
 

An example of graphing:f(x) = cos3x from -pi to pi

Here is how I would go about graphing the function f(x) = cos3x from -pi to pi, using our grid with marks at multiples of pi/12. Make sure your calculator is in radians mode before you begin.

cos(3×0) = 1

cos(3×pi/12) = sqrt(2)/2 ~= .7071
cos(3×2pi/12) =0
cos(3×3pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×4pi/12) = -1
cos(3×5pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×6pi/12) =0
cos(3×7pi/12) = sqrt(2)/2 ~= .7071
cos(3×8pi/12) = 1
cos(3×9pi/12) = sqrt(2)/2 ~= .7071
cos(3×10pi/12) =0
cos(3×11pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×12pi/12) = -1

Because cosine is an even function, the negative side of the graph is a mirror image of the positive side and I can use this information to make make work easier.

cos(3×-pi/12) = sqrt(2)/2 ~= .7071
cos(3×-2pi/12) =0
cos(3×-3pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×-4pi/12) = -1
cos(3×-5pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×-6pi/12) =0
cos(3×-7pi/12) = sqrt(2)/2 ~= .7071
cos(3×-8pi/12) = 1
cos(3×-9pi/12) = sqrt(2)/2 ~= .7071
cos(3×-10pi/12) =0
cos(3×-11pi/12) = -sqrt(2)/2 ~= -.7071
cos(3×-12pi/12) = -1

From those points, I know the graph of a sine or cosine function should be smooth instead of sharp, so I draw nice round curves going up to 1 and down to -1. Then, I draw the graph between sqrt(2)/2 and -sqrt(2)/2, making sure to pass through 0. As with everything I draw, it's never perfect, but it will give you a basic idea of what I expect on the homework and also on the next quiz. (Really big hint.)

Notice that the graph starts at its lowest point when x = -pi and repeats exactly 3 times. Whenever a graph is of the form f(x) = coskx or g(x) = sinkx, the distance from left to right in which the function repeats changes from 2pi to 2pi/|k|. If |k| > 1, this means the function repeats more quickly. If |k| < 1, this means the function repeat more slowly.

Practice with degrees and radians.

Here is a link to a blog post explaining degrees and radians with some practice problems.

Here are more practice problems for writing angles both in degrees and radians, where the degrees number is between 0° and 360° and the radians number is between 0 and 2pi.

This text editor can't write the superscript -1, so I'll use the "arc" prefix to discuss the inverse trig functions.

arccos(-.25): In degrees mode, the answer is 104.4775122...°, which I would round to 104.4775°. Changing to radians mode, the answer is 1.823476582..., which I would round to 1.8235 radians.
Dividing that answer by pi, I get .580430623..., which I would round to .5804pi.

arcsin(-.25): In degrees mode, the answer is -14.47751219...°. Adding 360° and rounding, the answer is 345.5225°.
Instead of changing to radians mode, I could just divide this by 180 to get 1.919569377... which is to say the reading rounds 1.9196pi radians.
Multiplying by pi, I get 6.030505052..., or 6.0305 radians.

Practice

a) arctan(-.25)

b) arccos(1/12)

c) arcsin(-1/12)

Answers in the comments.

Law of Cosines for third side length of a triangle, Law of Sines for area of the triangle

If we know the length of two sides of a triangle, call them a and b, and we have some way to find the angle between them, call it gamma, here is how we can use the Law of Cosines and the Law of Sines best. (Here we assume gamma is an angle between 0° and 180°.)

The third side length can be found using

c² = a² + b² - 2abcosgamma

and the area is given by

Area = ½absingamma

We need a and b as  numbers, but we can know gamma a few different ways.


1) gamma is given. Then we go to the calculator and find cosgamma and singamma.

2) cosgamma is given. We can find sine using The Trigonometric Identity. If we are asked for the measure of gamma, use the inverse cosine function.

3) singamma is given and the quadrant for gamma is given or we are told gamma is acute or obtuse.
We can find cosine using The Trigonometric Identity, where cosine will be positive if we are in the first quadrant (gamma is acute) or cosine will be negative if in the second quadrant (gamma is obtuse). If we are asked for the measure of gamma, use the inverse cosine function.


Example 1.

gamma = 135°, a = 6, b = 2.
sin135° = sqrt(2)/2, cos135° = -sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(-sqrt(2)/2) = 40 + 12sqrt(2), so c = sqrt(40 + 12sqrt(2)) or approximately 7.5479 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.


Example 2.

gamma = 45°, a = 6, b = 2.
sin45° = sqrt(2)/2, cos45° = sqrt(2)/2

c² = 6² + 2² - 2(6)(2)(sqrt(2)/2) = 40 - 12sqrt(2), so c = sqrt(40 - 12sqrt(2)) or approximately 4.7989 units.

Area =  ½(6)(2)sqrt(2)/2 = 3sqrt(2) or approximately 4.246 square units.

(Notice the area for the first two examples is the same.)


Example 3.

singamma = .2, gamma in first quadrant, a = 6, b = 2.

We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(2sqrt(6)/5) = 40 - 48sqrt(6)/5, so c = sqrt(40 - 48sqrt(6)/5) or approximately 4.0602 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.





Example 4.

singamma = .2, gamma in second quadrant, a = 6, b = 2.


We can also write singamma as 1/5, so 1/5² + cos²gamma = 1
1/25 + cos²gamma = 1
cos²gamma = 24/25
cosgamma = -sqrt(24/25) = -2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-2sqrt(6)/5) = 40 + 48sqrt(6)/5, so c = sqrt(40 + 48sqrt(6)/5) or approximately 7.9696 units.

Area =  ½(6)(2).2 = 1.2 square units exactly.
(Notice the area for examples 3 and 4 is the same.)

Example 5.

cosgamma = .2, a = 6, b = 2.

We can also write cosgamma as 1/5, so 1/5² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(.2) = 40 - 4.8, so c = sqrt(35.2) or approximately 5.9330 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.


Example 6.

cosgamma = -.2, a = 6, b = 2.


We can also write cosgamma as -1/5, so (-1/5)² + sin²gamma = 1
1/25 + sin²gamma = 1
sin²gamma = 24/25
singamma = sqrt(24/25) = 2sqrt(6)/5

c² = 6² + 2² - 2(6)(2)(-.2) = 40 + 4.8, so c = sqrt(44.8) or approximately 6.6933 units.

Area =  ½(6)(2)2sqrt(6)/5 or approximately 5.8788 square units.

Yet again, the areas of examples 5 and 6 are the same.


Note: this was updated on Oct. 11 to correct errors in the lengths of the third side. Thanks to my student Fred Johnson for pointing out the errors.

Answers to homework 5

Reading left to right, top to bottom

cos120° = -1/2 ||| tan225° = 1 ||| sin330° = -1/2 ||| cot240° = sqrt(3)/3


csc150° = 2 ||| sec210° = -2sqrt(3)/3 ||| sin135° = sqrt(2)/2 ||| tan300° = -sqrt(3)


cot315° = -1 ||| sin225° = -sqrt(2)/2 ||| sec270° = not defined ||| csc240° = -2sqrt(3)/3


a = 2, b = 3, gamma = 30°
Answer: c = sqrt(13 - 6sqrt(3)) ~= 1.6148
 a = 2, b = 3, gamma = 45°
Answer: c = sqrt(13 - 6sqrt(2)) ~= 2.1248

a = 2, b = 3, gamma = 60°
Answer: c = sqrt(7) ~= 2.6458

a = 2, b = 3, gamma = 90°
Answer: c = sqrt(13) ~= 3.6056

a = 2, b = 3, gamma = 120°
Answer: c = sqrt(19) ~= 4.3589

a = 2, b = 3, gamma = 135°
Answer: c = sqrt(13 + 6sqrt(2)) ~= 4.6352

a = 2, b = 3, gamma = 150°
Answer: c = sqrt(13 + 6sqrt(3)) ~= 4.8366

 

Three sides of a triangle:Area from Heron's formula, angles from the Law of Cosines

The Law of Cosine has three different statements, given side lengths a, b and c and angles opposite alpha, beta and gamma, respectively.

a² = b² + c² - 2bccosalpha
b² = a² + c² - 2accosbeta
c² = a² + b² - 2abcosgamma

On earlier homework, we figured out the third side from two lengths and the measure of the angle between them.  With a little algebraic manipulation, we can figure out the cosines of the three angles is we are given the three side lengths, and using the cosine inverse function, written here as arccos(number), we can find the three angles, or at least approximations, since the angles will often be given as irrational numbers.

cosalpha = [b² + c² - a²]/[2bc]
cosbeta = [a² + c² - b²]/[2ac]
cosgamma = [a² + b² - c²]/[2ab]

Examples

If we know the perimeter of a triangle is 7 and the sides are all whole numbers, there are only two possible answers, <3, 3, 1> and <3, 2, 2>. Let's answer all the questions from the quiz base on these numbers

c = 3, b = 3, a = 1
==============
perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and acute.

Area = sqrt(7/2[7/2 - 3][7/2 - 3][7/2 - 1]) = sqrt(7/2 × 1/2 × 1/2 × 5/2) = sqrt(35/16) = sqrt(35)/4
approximation: 1.4790

Because c = b, gamma = beta, so we only need to do one calculation to find both.
cosgamma = [1 + 9 - 9]/[2 × 1 × 3] = 1/6
arccos(1/6) ~= 80.4059°
So gamma = beta ~= 80.4059°

If we calculate alpha, we get cosalpha = [9 + 9 - 1]/[2 × 3 × 3] = 17/18
arccos[17/18] = 19.1881°
So alpha ~= 19.1881°

You get the same answer if you subtract 2gamma from 180°.

--
c = 3, b = 2, a =2
==============
perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and obtuse.

Area = sqrt(7/2[7/2 - 3][7/2 - 2][7/2 - 2]) = sqrt(7/2 × 1/2 × 3/2 × 3/2) = sqrt(63/16) = 3sqrt(7)/4
approximation: 1.9843

Because a = b, alpha = beta, so we only need to do one calculation to find both.
cosalpha = [9 + 4 - 4]/[2 × 3 × 2] = 9/12 = 3/4
arccos(1/6) ~= 80.4059°
So alpha = beta ~= 41.4096°

If we calculate gamma, we get cosgamma = [4 + 4 - 9]/[2 × 2 × 2] = -1/8
arccos[-1/8] = 19.1881°
So gamma ~= 97.1808°

You get the same answer if you subtract 2alpha from 180°.

Practice problems posted on Wednesday.

Find the classifications, exact and approximate area and approximate all three angles of the following triangles with perimeter = 11.

a) side lengths c = 5, b = 5, a = 1
b) side lengths c = 5, b = 4, a = 2
c) side lengths c = 5, b = 3, a = 3

Answers in the comments.


 

Link to a post about the Law of Cosines and some practice problems

Here is a post about Law of Cosines from 2011.

Here are examples with two side lengths a and b and the measure of between them, gamma, in degrees.  Use these to find the length of c exactly and rounded to four places after the decimal.

a = 3, b = 4, gamma = 45°

a = 3, b = 4, gamma = 90°

a = 3, b = 4, gamma = 135°

Answers in the comments.

The trig functions at perpendicular angles and at the opposite side of the circle (the antipode).

If you know the values of sine and cosine at some angle, call it alpha, you also know all the other trig values at that angle

tangent (a.k.a tan) = sine/cosine
cotangent (a.k.a cot) = cosine/sine
secant (a.k.a sec) = 1/cosine
cosecant (a.k.a csc) = 1/sine

More than that, we know the values of sine and cosine at alpha+90°, alpha+180° and alpha+270°. Here are the rules for sine and cosine, and from those all the rest fall into place.

perpendicular clockwise
sin(alpha+90°) = cos(alpha)
cos(alpha+90°) = -sin(alpha)

antipodal (going the opposite direction on the same line)
sin(alpha+180°) = -sin(alpha)
cos(alpha+180°) = -cos(alpha)

perpendicular counter-clockwise
sin(alpha+270°) = -cos(alpha)
cos(alpha+270°) = sin(alpha)

These rules will answer all the questions about the trig values at the angles listed on this homework. Make sure to give the answers in exact values with no square roots in the denominators.

Practice for finding the other trig functions and the angle from the value of sine, cosine or tangent, assuming all are positive

Example #1: cosalpha = 2/5

(2/5)² + sin²alpha = 1
4/25 + sin²alpha = 1
sin²alpha = 21/25
sinalpha = sqrt(21)/5

since we have sine and cosine, tan = sin/cos.

tanalpha = [sqrt(21)/5]/[2/5] = sqrt(21)/2

With this editor, I can't write -1 as a superscript, so instead I will use the words arcsine, arccosine and arc tantangent. On your calculator, arccos(2/5) = 66.42182152...°, which we round to 66.4218°.

If you take the unrounded value and change it to degrees-minutes-seconds, rounding the seconds to the nearest whole number, we get 66° 25' 19".

Example #1.1: If we had sinalpha = 2/5, the work would look nearly identical, except cosalpha would equal sqrt(21)/5.  Tangent of this angle is the reciprocal of sqrt(21)/2, which is 2sqrt(21)/21 when written in rational denominator form.  The angle is the complement of our original angle, which means they add up to 90°. arcsin(2/5) = 23.57817848...°, which rounds to 23.5782°.

The DMS version of the unrounded value, rounded to the nearest whole second is 23° 34' 41".

Example #2: tanbeta = 2/5

Since tan = sin/cos,
tan × cos = sin

Using this, cos²beta + (2/5)²cos²beta = 1

cos²beta + 4/25cos²beta = 1
29/25cos²beta = 1
cos²beta =25/29
cosbeta = sqrt(25/29) = 5/sqrt(29) = 5sqrt(29)/29

arctan(2/5) = 21.80140949...°, which rounds to 21.8014°, and in DMS is 21° 48' 5".

Here are practice problems. Get the other two trig values, the angle rounded to the nearest ten thousandth of a degree and rounded to the nearest second in DMS mode.

1) cosalpha = 1/10

2) tanbeta = 1/10

3) singamma = 1/9

4) tandelta = 1/9

Answers in the comments.

The exact trig values "famous" angles from 0° to 90°

Here is a handy reference for the major trigonometric function values for the "famous" angles in the first quadrant and at 0° and 90°, which technically are on the x-axis and y-axis respectively, not in the first quadrant but instead on the borders.

Here are a few well known properties of these functions over this range.

1. Sine and tangent are increasing.  While sine increases, it is bounded by 1 as the highest possible value. Tangent is unbounded and can get as large as any positive number you can name.  As I said in class, tangent is the same as the slope of the line that connects the origin (0, 0) to the point (cos alpha, sin alpha). (The editor in Blogger is bound by the symbols available in HTML, so I will have to spell out Greek letters when used as variables.)

2. Cosine is decreasing over the range.

3. Cosine is short for "complementary sine", which means cos(90°-alpha)= sin alpha, when alpha is an angle measured in degrees. In the other direction, sin(90°-alpha)= cos alpha.

Link to an explanation of The Trigonometric Identity

There are many trigonometric identities, but except for some definitions of terms, nearly all of them are re-arrangements of what I call The Trigonometric Identity

sin²alpha + cos²alpha = 1

Here is a link to the post from last year.

Euclid's proof and similar triangles

Using the handout from today, we know Euclid's proof is based on similar triangles.  With two similar triangles, the ratio between corresponding sides remains constant. If we list the sides of each triangle in the order short-leg-long leg-hypotenuse, here are the three triangles.

The big triangle: a-b-c
The triangle with b as hypotenuse: h-y-b
The triangle with a as hypotenuse: x-h-a

Using this information, given the lengths of one triangle we can find all the others.

Example: a-b-c are 3-4-5

The triangle with b as hypotenuse: h-y-b
h/3 = y/4 = 4/5

h = 12/5, y = 16/5


The triangle with a as hypotenuse: x-h-a
x/3 = h/4 = 3/5

x = 9/5, h = 12/5

Not surprisingly, we get the same value for h in both similar triangles it belongs to.

practice problems:

a) h-y-b are 3-4-5

b) x-h-a are 3-4-5

c) a-b-c are 8-15-17

Answers in the comments.

Slope, distance and the area of a triangle defined by three points

Here is a link to practice problems for slope, distance and the area of a triangle defined by three points.

Here are another three points and the origin: (1, 7) (4, 6) (3,-3) and (0,0)

Slope problems.
Find the slope of the line connecting the points. Write them as fractions in lowest terms. (Note: if the answer is a whole number, leave it as a whole number. 3 is simpler than 3/1.)

a) (1,7) and (4,6)
b) (1,7) and (3,-3)
c) (1,7) and (0,0)
d) (4,6) and (3,-3)
e) (4,6) and (0,0)
f) (3,-3) and (0,0)

Distance problems
Find the distance between the pairs of points and write the answer as a simplified square root.
a) (1,7) and (4,6)
b) (1,7) and (3,-3)
c) (1,7) and (0,0)
d) (4,6) and (3,-3)
e) (4,6) and (0,0)
f) (3,-3) and (0,0)

Areas of triangles
Find the areas of the triangles defined by the following three points.
a) (1,7), (4,6) and (0,0)
b) (1,7), (3,-3) and (0,0)
c) (4,6), (3,-3) and (0,0)
d) (1,7), (4,6) and(3,-3)

Answers in the comments.

Proofs of Heron's formula and the ½|ad - bc| area method

After class on Thursday, a student asked for proofs of our two methods for getting the area of a triangle.

Here is a link to the Wikipedia page on Heron's formula which has several proofs.

For the triangle, here is a page proving |ad - bc| is the area of a parallelogram, which is done by cutting a parallelogram out of a larger rectangle. Any parallelogram can thought of as two congruent triangles glued together along one corresponding side, so half the area of a parallelogram is the area of one of the triangles.

Links to material covered in first week and on first homework.

The blog has entries from fall 2011, the last time I taught this course.  I expect to follow much the same order of material this time around.  Here are links to the topics we worked on this first week.

Classifications of triangles and The Triangle Inequality.

Practice on classification of triangles given two angles.

Practice on classification given three side lengths.

In class on Thursday, I asked for all possible integer length side triples for perimeter of 11 and perimeter of 13.  Here are the answers.

11
5, 5, 1
5, 4, 2
5, 3, 3
4, 4, 3

13
6, 6, 1
6, 5, 2
6, 4, 3
5, 5, 3
5, 4, 4

Find both classifications for these triples, the classification by largest angle (obtuse, right, acute) and the classification by relations between side lengths (equilateral, isosceles, scalene).

Answers in the comments.

Syllabus for Fall 2012

Math 50: Trigonometry (L-41147)            Fall 2012 – Laney College
Instructor:  Matthew Hubbard                TTh: 10:30-11:45 am
Email address: mhubbard@peralta.edu       
Recommended Text: Trigonometry (open source) http://mecmath.net/trig/trigbook.pdf
Class website: http://thetrig.blogspot.com/

Office hours:     MW: 8:30-8:55 am F-203 (Classroom)
        T-Th: 10:00-10:25 am G-206 (Classroom)

Add and drop class dates
Last date to add:                    Sun., Sept. 2
Last date to drop class without a “W”:        Sat., Sept. 1
Last date to drop class with a “W”:            Sat., Nov. 17

Holiday schedule for T-Th schedule
Thanksgiving                        Thursday, Nov. 22

Test dates:
Midterm 1:        Thursday, Sept. 27
Midterm 2:        Thursday, Nov. 1
Comprehensive Final:    Thursday, Dec. 13 10:00 am- noon

    Homework to be turned in:  Assigned last class period of the week, due next class.
    Late homework accepted AT THE BEGINNING of the class after it was due
    Quizzes:    One on the last class of the week in weeks without a midterm or final

        Grading system

Homework            20%
Lab                  5%
Quizzes            25%
Midterm I            25%
Midterm II            25%
Final                25%
Voting bonus              2%   
 If you are eligible to vote, bring an “I voted” sticker
                    If you are too young to vote, bring proof of age
                    If you are not a U.S. citizen, bring proof of status 

Lowest two of the homework scores will be dropped from the total.
Lowest two of the quiz scores will be dropped from the total.
Lowest total out of 100 points the quiz total and two midterms will be dropped from the final grade.
Anyone getting a higher grade out of 100 points on the final than the weighted average of all grades combined will get the final percentage instead deciding the final grade.  This option is only available to students who have missed at most two homework assignments.
Anyone going into final with 97.5% class average can skip the final and be awarded an A.


Class rules:  All cell phones and electronic communication devices off during class.
No hats, hoodies or headphones worn during quizzes and exams.
No calculators that also combine a cell phone or text message machine.

Recommended calculator: TI-30XIIs (any calculator with at least two lines of output will do, the TI-30XIIs is the cheapest that does all the things you need to do in this class.  If you need help with any Texas Instruments calculator, I should be able to steer you in the right direction.  I haven’t used other brands of calculators as much.)

Open source textbook:  The textbook is free and online.  It helps a lot if you have a computer and Internet at home so you can have your own copy on your computer.  You will need the Adobe Reader software, which is free and can be downloaded from several websites.  I will be assigning problems from the book occasionally.

Academic honesty:  All assignments you turn in, homework, exams and quizzes, must be your own work.  Anyone caught cheating on these assignments will be punished, where the punishment can be as severe as failing the class or being put on college wide academic probation.

Students with disabilities
The Disabled Students Program Services (DSPS) should have your academic accommodation with the instructor.  After the first day, I will accept these accommodations electronically or by hard copy on paper.  If you need academic accommodation and have not yet applied, please call 510-464-3428 for an appointment.

Exam policies
Tests will be closed book and closed notes. Some information you will be expected to remember, other formulas and information will be provided. No sharing of calculators is allowed.  You are responsible for knowing how to use your calculator to find answers.

            Student learning outcomes

Math 50 Trigonometry
•    Evaluate the 6 trigonometric functions using a calculator, as well as determining exact values for some special angles without a calculator.
•    Solve a triangle (right, acute, obtuse), given various angles and sides.
•    Convert between decimal degrees, degree-minute-seconds, and radian measure of an angle.
•    Demonstrate knowledge of several trigonometric identities and use them to verify other identities.
•    Graph trigonometric functions.
•    Solve trigonometric equations.

The reciprocal relationship

The teacher will be on time and prepared to teach the class.
The students will be on time and prepared to learn.

The teacher will present the material to the best of his ability.
The students will absorb the material to the best of their ability.  They will ask questions when topics are not clear.

The teacher will do his best to answer the questions the students ask about the material, either by repeating an answer with more details included or by taking a different approach to the material that might be clearer to some students.
The students will understand if the teacher feels a topic has been covered enough for the majority of the class and will accept questions being answered outside the class, either in extra time or through written communication.

The teacher will do his best to keep the class about the material.  Personal details and distractions that are not germane to the class should not be part of the class.
The students will do their best to keep the class about the material.  Questions that are not about the topic should be avoided.  Distractions like cell phones and texting are not welcome when the class is in session.

The teacher will give assignments that will help the students master the skills required to pass the course.
The students will put in their best efforts to complete the assignments.
When the assignments are completed, the teacher will make every effort to get the assignments graded and back to the students in a timely manner, by the next class session whenever possible.

The teacher will present real life situations where the skills being learned will be used when they exist.  In math, sometimes a particular skill is needed in general to solve later problems that will have real life applications.  Other skills have the application of “learning how to learn”, of committing an idea to memory so that committing other ideas to memory becomes easier in the long run.
The student has the right to ask “When will I use this?” when dealing with mathematical topics.  Sometimes, the answer is “We need this skill for the next skill we will learn.”  Other times, the answer is “We are learning how to learn.”  Both of these answers are as valid in their way as “We will need this to understand perspective” or “We use this to balance our checkbooks” or “Ratios can be used to figure out costs” or other real life applications.