Part 4, Set 2: Periodic functions & Trig Identities

The graphs of the sine and cosine functions look likes waves. In science and engineering, lots of things oscillate (wave up and down), and the sine and cosine functions are very useful in working with these oscillations. Examples include sound waves, radio waves, light waves, and seismic waves. Even different shaped waves that repeat can be analyzed using the sine and cosine functions (this is called spectral analysis).

Periodic Functions

There are three main ways that one wave can be different from another: a) starting point: example sine versus cosine;b) how big the wave is: amplitude;c) how quickly in time the waves oscillate: period (or its inverse, frequency), or how long a distance it takes the wave to go up and down (wavelength).

Periodic Functions

How do sine and cosine waves differ? Only in the starting point. Can we have waves that start somewhere inbetween? If so, how do we describe that starting point?

Periodic Functions

We saw before that sin(θ+90o) = cos(θ) . From this we see that a cosine function is simply a sine function that starts out at an angle of 90o instead of starting out at 0o. We can generalize this by replacing the 90o with an initial angle, θo.

Our sine function now looks like y(θ) = sin(θ + θo).

Periodic Functions

Next, the sine function has a range of [-1,1]. In science and engineering, waves have heights or magnitudes. We take this into account by introducing an amplitude, A, which multiplies the sine function. The amplitude will include any units that the quantity has (for example, height or energy).

Our sine function now looks like y(θ) = A*sin(θ + θo).

Periodic Functions

Periodic functions usually depend on time. The sine function operates on angles. How do we work this situation out?

Periodic Functions

All we need to do is convert the time into an angle by a conversion coefficient. In physics we define the term angular velocity with the symbol, ω, to be: ω = Δθ/Δt . For many cases, ω will be a constant. In those cases, we also define the term, period, with the symbol, T, to be the time it takes the angle to go through a complete cycle = 2π radians. Thus we have ω = 2π/T. Our sine function now looks like: y(t) = A*sin(ω*t + θo), or assuming the *: y(t) = A sin(ωt + θo) = A sin(2πt/T + θo) .

Periodic Functions

In some cases, the oscillating function oscillates over distance instead of over time. We do the same kind of thing, though in this case we use a constant that converts the distance into an angle: k = Δθ/Δx, called a wave number. For those cases where k is a constant, we define a distance over which the angle changes by 2π and call that distance the wavelength, λ. We now have k = 2π/λ, and our oscillating function looks like: y(x) = A sin(kx + θo) = A sin(2πx/λ + θo) .

Periodic Functions

Note that the angles used in periodic functions do NOT refer to geometric angles, but rather to specific points on the sine curve and these angles are called phase angles. We do not usually measure these angles with protractors or other measuring devices directly, and so these phase angles are almost exclusively expressed in radians rather than degrees.

Periodic Functions

When the phase angle of a sine wave is at π/2 radians = 90o, the sine is at a maximum and this is called the crest of the wave. When the sine wave is at 3π/2 radians = 270o, the sine is at its lowest value and this is called the trough of the wave.

Composite Functionan example

A nice example of a composite function (see Part 1-2, on or near slide 53) happens with periodic functions. The sine function depends on the phase angle, θ; and that phase angle depends on time (and/or distance):

f(θ) = A sin(θ), and g(t) = θ(t) = ω*t + θo

Recall that the notation for this: (f ◦ g)(x) = f(g(x)).

In our example here: f(θ(t)) = A sin(ω*t + θo).

Periodic Functions

We have seen that a sine function is periodic, and it repeats over a phase angle of 2π radians.

We have seen that a cosine wave is simply a sine wave that has been phase shifted by π/2 radians or 90o.

The tangent function is also periodic, but it repeats over an angle of π radians (180o) rather than 2π radians (360o) like the sine and cosine (see at or near slide #42 in the previous set, or graph the tangent function on your calculator, or look at the definition of tangent).

Periodic Functions

There are three other periodic functions that have been defined. These are just the multiplicative inverses of our three main periodic functions of sine, cosine, and tangent:

cotangent: cot(θ) = 1 / tan(θ)cosecant: csc(θ) = 1 / sin(θ)secant: sec(θ) = 1 / cos(θ)

Periodic Functions

cotangent: cot(θ) = 1 / tan(θ)cosecant: csc(θ) = 1 / sin(θ)secant: sec(θ) = 1 / cos(θ)Note: since sine, cosine, and tangent all have

angles where they are zero, all three of the above functions “blow up” (approach infinity / are undefined) at those phase angles and so have asymptotes there.

Trig Identities

There are many different ways you can write the Pythagorean Theorem: x2 + y2 = r2.We have already looked at this in this form:sin2(θ) + cos2(θ) = 1 . Using the functions we defined on the previous slide, we have:1/csc2(θ) + 1/sec2(θ) = 1. We could multiply through by csc2(θ) to get:1 + csc2(θ)/sec2(θ) = csc2(θ). If we recognize that csc(θ)/sec(θ) = cos(θ)/sin(θ) = cot(θ), we could rewrite the previous one to be:1 + cot2(θ) = csc2(θ).

Trig Identities

There are several more ways of re-writing the Pythagorean Theorem. Can you come up with at least one more way?

Can you get a qualitative idea of what the graphs of the sec(θ), csc(θ), and cot(θ) look like? Since each is a multiplicative inverse of cos(θ), sin(θ), and tan(θ) respectively, each will have the same sign as its partner, and when its partner is small it will be big, and when its partner is big it will be small. Note any symmetries and where any asymptotes should be.

Trig Identities

Because we sometimes have periodic functions interacting in various ways, we sometimes get weird expressions that would be easier for us to recognize the behavior if we could simplify them. Usually, you just convert the last three (sec, csc, & cot) to their multiplicative inverses (cos, sin, & tan), and then use the Pythagorean Theorem. The text shows several examples.

More Useful Trig Identities

Sometimes we have different terms inside a periodic function (like sine or cosine). It is often useful to be aware of (remember or know to look up) these four (two pair):

sin(θ±φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)cos(θ±φ) = cos(θ)cos(φ) sin(θ)sin(φ)Note the on the second identity is opposite

the ± on the first one.

More Useful Trig Identitiessin(θ±φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)cos(θ±φ) = cos(θ)cos(φ) sin(θ)sin(φ)I don’t always remember which is which, but I know

how to test which is which: if both θ & φ are 0o = 0 radians, then sin(θ+φ) = 0 and the top expression sin(θ)cos(φ) ± cos(θ)sin(φ) = 0 but the bottom expression cos(θ)cos(φ) sin(θ)sin(φ) = 1.

To determine how the ± or goes (regular or up-side-down), if both θ & φ are 45o = π/4 radians, then sin(θ+φ) = 1 and sin(θ-φ) = 0 and see whether the ± or the sign works to give 0 or 1.

More Useful Trig Identities

Since there are identities for sin(θ±φ) and cos (θ±φ), we might expect there to be one for tangent also, and here it is:

tan(θ±φ) = [tan(θ) ± tan (φ)]/[1 tan(θ)tan(φ)]

More Useful Trig Identities

Example: AM radio works by having a carrier way of high frequency and then it modulates the amplitude (hence AM). The wave looks like: [A sin(ω1t)]*sin(ω2t) . By using the trig identities on the previous slide, this becomes upon adding the two below -cos(θ+φ) = -cos(θ)cos(φ) - -sin(θ)sin(φ) cos(θ-φ) = cos(θ)cos(φ) + sin(θ)sin(φ)

cos(θ-φ) – cos(θ+φ) = 2 sin(θ)sin(φ), or

A sin(ω1t)sin(ω2t) = ½ A cos(ω1t - ω2t) – ½ A cos(ω1t + ω2t) =A sin(ω1t)sin(ω2t) = ½ A cos([ω1- ω2]t) – ½ A cos([ω1+ω2]t)

More Useful Trig IdentitiesUseful special cases of these trig identities are

the double angle identities where we let θ=φ:sin(θ+φ) = sin(θ)cos(φ) + cos(θ)sin(φ)sin(2θ) = sin(θ)cos(θ) + cos(θ)sin(θ) , or sin(2θ) = 2 sin(θ) cos(θ) , and cos(θ±φ) = cos(θ)cos(φ) sin(θ)sin(φ) cos(2θ) = cos(θ)cos(θ) - sin(θ)sin(θ) , or cos(2θ) = cos2(θ) – sin2(θ) , andtan(2θ) = sin(2θ)/cos(2θ) = 2 sin(θ) cos(θ) /[cos2(θ) – sin2(θ)] , divide thru by cos2(θ) gives: tan(2θ) = 2tan(θ)/[1-tan2(θ)]

More Useful Trig Identities

We can convert the double angle identities into half angle identities: cos(2θ) = cos2(θ) – sin2(θ) using the Pythagorean Theorem: 1 = sin2(θ) + cos2(θ)cos(2θ) = cos2(θ) – [1 – cos2(θ)] = 2cos2(θ)-1,or cos2(θ) = ½ [1+cos(2θ)], orcos(θ/2) = ±SQRT[½{1+cos(θ)}]

More Useful Trig Identities

We again start with cos(2θ) = cos2(θ) – sin2(θ) and use the Pythagorean Theorem: 1 = sin2(θ) + cos2(θ) to get cos(2θ) = [1 – sin2(θ)] - sin2(θ) = 1 - 2sin2(θ),or sin2(θ) = ½ [1-cos(2θ)], orsin(θ/2) = ±SQRT[½{1-cos(θ)}] .

Using the definition of tangent as sine / cosine:tan(θ/2) = sin(θ/2)/cos(θ/2) = ±SQRT[½{1-cos(θ)}] / ±SQRT[½{1+cos(θ)}] = tan(θ/2) = SQRT[{1-cos(θ)} / {1+cos(θ)}]

Other triangles

There are two more laws or relationships that are useful when we do NOT have right triangles: the law of cosines and the law of sines.

Other trianglesConsider the triangle below (solid lines) and the

extensions to make it into a right triangle. From the Pythagorean Theorem, we have: (a+d)2 + e2 = b2 , or a2 + 2ad + d2 + e2 = b2; from the extended triangle (with the dotted lines), d2 + e2 = c2. This simplifies our expression toa2 + 2ad + c2 = b2 . Use cos(θ) = (a+d)/b, or d = b*cos(θ) – a. Using this in our previous eq.:a2 + 2a[b*cos(θ) – a] + c2 = b2 which gives a2 + 2ab*cos(θ) – 2a2 + c2 = b2 or 2ab*cos(θ) = a2 + b2 – c2

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b

θ

Other triangles2ab*cos(θ) = a2 + b2 – c2

This is called the law of cosines. Note that a and b are the sides next to the angle, θ, and c is the side opposite the angle.

Note thatif θ < 90o = π/2 radians then cos(θ)>0 and c2 < a2 + b2; if θ = 90o = π/2 radians then cos(θ)=0 and c2 = a2 + b2;if θ > 90o = π/2 radians then cos(θ)<0 and c2 > a2 + b2.Can you see this in the diagram on the previous slide?

Other trianglesAgain consider the triangle below (solid lines) and the

extensions to make it into a right triangle, and note that β = 180o–φ. Note that: sin(θ) = e/b, and that the sin(β) = e/c = sin(180o–φ). If I add 180o to β, sin(180o+β) = -sin(β) = sin(180o+180o–φ) = sin(–φ) = -sin(φ), so that sin(β) = sin(φ), and sin(φ) = e/c.We now re-write the purple expressions as:e = b*sin(θ) and e = c* sin(φ). Since both are equal to e, both are equal to eachother: b*sin(θ) = c* sin(φ), or sin(θ)/c = sin(φ)/b.

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b

θβ φ

Other trianglessin(θ)/c = sin(φ)/b

Note the side c is opposite the angle θ, and the side b is opposite the angle φ. This is true for any two angles and sides, and so is true for all three angles and sides. This law is known as the law of sines.

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b

θβ φ

Regular Homework Set #9(continued on the next slides)

1. Choose three angles in at least two different quadrants and show for each that sin(θ+90o) = cos(θ) .

2. Given the following oscillation formula: h = (8 cm) sin([23 rad/sec]t + π/4 rad)find the period, T, of the oscillation and its frequency. [Frequency is defined as 1/T.]

3. Express each of these angles in degrees and find the sine, cosine, tangent and secant for each:a) π/5 rads; b) 2π/7 rads; c) 1.97 rads.

Regular Homework Set #9(continued from the previous slide)

4. Create another form (in addition to the ones on or near slide #14) for the Pythagorean Theorem.

5. Hand draw the graph of sec(θ).6. For a (non-right) triangle with side =3 cm, side =

6 cm, and angle inbetween = 120o, find the connecting side length and the other two angles.

7. For a (non-right) triangle with side = 6 cm and the angle on one side = 120o and the other angle 45o, determine the other two sides and the other angle.

Regular Homework Set #9(continued from the previous slide)

8. Two people that are a horizontal 20 feet apart both site an object in the distance. The angle that the object makes with the connecting line between thetwo people is 80o for oneand 98o for the other – see the diagram. How far away is the object from the nearer person?

80o

20 ft

98o

x

Regular Homework Set #9(continued from the previous slide)

9. Two people that are a horizontal 20 feet apart both site an object in the sky. The angle that the object makes with the connecting line between thetwo people is 80o for oneand 98o for the other – see the diagram. How high is the object?

80o

20 ft

98o

h

WebCT Tests #4-7

You should be able to do the fourth, fifth, sixth and seventh WebCT tests now.