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.


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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.