Polar coordinates.

In class Monday, we introduced the idea of polar coordinates.  In rectilinear coordinates (x, y), the first number tells us how far right (positive) or left (negative) we move on the horizontal axis and the second number tells us how far up (positive) or down (negative) we move on the vertical axis.

In polar coordinates we have (r, theta).  There is a central point, not unlike (0, 0) in the xy-axis system, and we consider that we are pointing in a direction we call angle 0 in radians.  The standard is to make that direction to the right, the same as the positive x-axis in rectilinear.  r is the distance from the origin and theta is the angle given in radians.

It is acceptable to have negative values for distance and for angles.

Unlike rectilinear coordinates, polar coordinates are not unique.  The easiest example of this are the coordinate pairs (0, 0) and (0, 1).  The first zero means we are a distance of 0 from the origin.  That means we are on the origin.  The angle we turn doesn't change where we are.

Even when points aren't the origin, the polar coordinates are not unique. The simplest example is

(1, 0) = (1, 2pi)

What this says is if we are 1 away from the origin and pointing to the right, this is the same as being 1 away from the origin and turning a full circle. In fact, adding or subtracting any multiple of 2pi to an angle brings us back to pointing in the exact same direction.

We will be working more in polar coordinates for the rest of the week, possibly longer.

<br>

step by step through some half angles of famous angles.

The half angle formula deal with fractions and square roots.  The Blogger text software doesn't have good ways to type those in, so I resort to using pictures with captions.

These are the original half angle formulas.

This is a step by step (across) way to turn the sine and cosine of 15 degrees into fractions with whole number denominators.  These answers look different from the ones you get using subtraction of angles, either 60 degrees - 45 degrees or 45 degrees - 30 degrees.  If you type in the formulas, you get the same answers on your calculators.

This is a step by step (across) way to turn the sine and cosine of 22.5 degrees into fractions with whole number denominators. 

Answers to quiz 9

cosalpha = 1/5_____ sinalpha = 2sqrt(6)/5 _____ tanalpha = 2sqrt(6)

cosbeta = 2/5______ sinbeta = sqrt(21)/5 ______ tanbeta = sqrt(21)/2

cosgamma = 3/5___ singamma = 4/5 _________ tangamma = 4/3

cosdelta = 4/5 _____sindelta = 3/5 ___________ tandelta = 3/4 

cos(alpha + beta) = (2 - 6sqrt(14))/25
sin(alpha + beta) = (sqrt(21) + 4sqrt(6))/25
tan(alpha + beta) = 50(sqrt(21)+sqrt(6))/-500 = -(sqrt(21)+sqrt(6))/10

Corrected from original.  Thanks to Laura Hidrobo.

cos(beta - gamma) = (6 + 4sqrt(21))/25
sin(beta - gamma) = (3sqrt(21) - 8)/25
tan(beta - gamma) = 1 - sqrt(21)/6

Corrected from original.  Thanks to Bi Ting Zhen and Laura Hidrobo.

cos(gamma + delta) = 0
sin(gamma + delta) = 1

tan(gamma + delta) =undefined

The last one is this way because the angles are complementary.

ERROR ON HOMEWORK #12 Corrected in class, but just to make sure.

Cos 72° = (sqrt(5) - 1)/4

Cos 36° = (sqrt(5) + 1)/4

Sorry for the screw-up. See you on Monday.

Half angle formulas

Instead of typing out the half angle formulas, here is some artwork that spells them out nicely.  Tangent is sine divided by cosine, so you would expect the square root sign to be involved, but a little algebraic manipulation makes it go away, as we will see in class Wednesday.

It all depends on whether we multiply the fraction under the square root sign by (1+ cosa)/(1 + cosa) or (1 - cosa)/(1 - cosa).

Here are some practice problems.  Assume the angles are in the first quadrant, which means that half the angle must also be in the first quadrant, so all the major trig functions are positive.

a) cosa = 1/3

b) cosb = 1/4

c) cosc = 1/5

Answers in the comments.

Addition and subtraction of angles and the double angle formula

Addition of angles formulae:               
cos(a + b) = cosacosb   – sinasinb      
sin(a + b) = cosasinb   + sinacosb          

Subtraction of angles formulae:
cos(a – b) = cosacosb   + sinasinb
sin(a – b) = sinacosb  - cosasinb

The double angle formulae:
cos(2a) = cos²a - sin²a
sin(2a) = 2cosasina

For the tangents of the new angles, divide sine by cosine.

Consider two angles a and b, with the following sine and cosine values.

sina = 1/3 and cosa = 2sqrt(2)/3

sinb = 2/3 and cosb = sqrt(5)/3

Find the following values as exact numbers.

sin(a + b)
cos(a + b)
tan(a + b)


sin(a - b)
cos(a - b)
tan(a - b)



sin(2a)
cos(2a)
tan(2a)


Answers in the comments.


 

msinx + ncosx

Adding multiples of sinx and cosx creates a new sine wave that is taller and has a phase shift.  The rules are as follows:

f(x) = msinx + ncosx

Amplitude = sqrt(m² + n²) 

Phase shift: f(x) = 0 when
msinx = -ncosx

With a little algebraic manipulation we get f(x) = 0 when tanx = -n/m. Therefore the phase shift is -arctan(-n/m).  This is where the function will equal 0, but you have to check to see if it is rising at that point or falling.

If it is rising, we can re-write f(x) = sqrt(m² + n²)sin(x - arctan(-n/m)).

If it is falling, f(x) = -sqrt(m² + n²)sin(x - arctan(-n/m)).

I created this picture over at the website WolframAlpha by typing in "plot sinx + cosx and sinx + 2cosx and 2sinx + cosx", without the quotation marks.









This picture is of all the possible choices of +/-sinx +/- cosx.  The inverse tangent of +/- 1 is +/- pi/4, so those are the places where the graphs cross the x-axis closest to the origin.  The new amplitude for all of these is sqrt(1² + 1²) = sqrt(2).  These four examples tell us how the four options work

f(x) = sinx + cosx:  Graph is in blue.  It crosses the axis at x = -pi/4 and goes upwards so it can be re-written as
f(x) = sqrt(2)sin(x + pi/4)

g(x) = -sinx - cosx:  Graph is in green.  It crosses the axis at x =-pi/4 and goes downwards so it can be re-written as
g(x) = -sqrt(2)sin(x + pi/4)


h(x) = sinx - cosx:  Graph is in red.  It crosses the axis at x = pi/4 and goes upwards so it can be re-written as
h(x) = sqrt(2)sin(x - pi/4)

k(x) = -sinx + cosx:  Graph is in yellow.  It crosses the axis at x = pi/4 and goes downwards so it can be re-written as
k(x) = -sqrt(2)sin(x - pi/4)

Rules for the signs in front of the amplitude and the phase shift.

The sign in front of phase shift is positive if m and n agree in sign (either both positive or both negative) and is negative if m and n disagree in sign.

The sign in front of the amplitude agrees with the sign in front of the sine function.  (I incorrectly stated in class it was the sign in front of the cosine function.  Sorry for the error.)

We will have a lab to practice this on Wednesday.
<br>