Practice for Homework 1, due January 31

Heron's Formula practice #1

Heron's Formula practice #2 

Classification of triangles by angle 

Classification of triangles by side length

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

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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.