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## Integration: Sines, Cosines and U-Substitution

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

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## Find solutions to a Trig Function

Q:  Find all solutions for X where 0 ≤ X < 360 degrees and 9*cos(X) – 5 = 0.

Okay, the 0 ≤ X < 360 just tells you we are looking for answers in one complete circle (no more, no less).

So, we first need to solve for X like so:

9*cos(X) – 5 = 0

9*cos(X) = 5

cos(X) = 5/9

X = cos-1(5/9)

Plug this into your calculator and you get:

X = 56.25 degrees

This is part of your answer, but not all of it.  The calculator will only give one answer, and will not consider answers in multiple quadrants.  We will use this answer to find any other solutions for X.

From our problem, we saw that:

cos(X) = 5/9

this means that the cosine value is positive.  In what quadrants are the cosine values positive??

So, X = 56.25 degrees is the answer for the angle in Quadrant 1.

To get the angle in Quadrant 4 that also works, you gotta do 360 – 56.25 = 303.75

Therefore, 303.75 degrees is another answer for X.

What this is saying is that:

cos(56.25)= 5/9 and cos(303.75) = 5/9 also.

So, X = 56.25 and 303.75

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## Simplifying Trig Expressions Using Identities

Q:  Simplify cos4t – sin4t

A:

First thing we gotta notice is that this is a difference of squares.  cos4t is the same as (cos2t)2.  So, we need to use what we know about factoring from algebra to factor this:

cos4t – sin4t can be factored using the difference of squares formula/concept (this just takes practice to see this and realize it):

(cos2t – sin2t)(cos2t + sin2t)

But, now we can use a trig identity because cos2t + sin2t = 1, so plug that in:

(cos2t – sin2t)(cos2t + sin2t)

(cos2t – sin2t)(1) = (cos2t – sin2t)

Now we gotta notice that even though it is simplified a bunch, we have another identity:

(cos2t – sin2t) = cos(2t)

And now we are as simplified as we get.

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## Amplitude, Phase Shift and Period

Q:  Consider the function y = -5 cos (2x – .1π)

a)  Identify the amplitude

b)  Identify the phase shift

c)  Identify the period

The following formulas / concepts will work for sin and cos graphs:

y = a cos (b*(x – h) ) + k

Note:  a, b, h and k are just numbers that will affect the graph.

The amplitude is |a|

The phase shift if 2π / |b|

The phase shift is “h”

The vertical shift is “k”

A big thing to notice:  the “b” value is factored out in this red formula.

So…. Let’s start with our actual example: y = -5 cos (2x – .1π)

Identify who is a, b, h, and k.  Notice, the b is not factored out, so let’s do that firt:

y = -5 cos (2x – .1π)

y = -5 cos (2*(x – .05π))

[if you multiply the 2 back through, you get the same as the original]

Now,

a = -5

b = 2

h = .05*pi

k = 0 (there is no + k at the end of the problem)

a)  The amplitude is |a| = |-5| = 5

b)  The phase shift is h = .05π (the graph is shifted .05π units to the right)

c)  The period is 2π/|b| = 2π/|2| = π

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## Trig Functions and Exact Values

Q:  Let X be an angle in quadrant III such that cos(X) = -12/13.  Find the exact values of csc(X) and cot(X).

A:

This will be a little hard to explain, since I cannot currently provide a picture, but it should be manageable!  Draw a picture on your own paper and follow along with me.

Draw a right triangle and label one of the angles X (just not the right angle).  We are going to ignore the negative sign for now and then take it in to account later!  Since the cos(X) = 12/13, the side adjacent to X is 12 and the hypotenuse is 13.  Use the Pythagorean Theorem to find the missing side (the side opposite X).  You will find that the opposite side is 5.

So, Adj = 12, Opp = 5 and Hyp = 13

Now, let’s consider the negative sign:  Since X is a QIII angle, we know that only tangent and cotangent are positive values in that quadrant — all other trig functions in QIII are negative.

Now, we need to find the csc(X).  “csc” is the ratio of the hyp / opp [and, it will be negative]

So, csc(X) = -13/5

And, cot(x) is the ratio of the adj / opp [and, it will be positive].  So, cot(X) = 12/5

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Q : Identify the x-values that are solutions of each multiple choice equation:

1.  8cos x +4 = 0

A. x= 2π/3 , 4π/3
B. x=4π/3, 5π/3
C. x = π/3 , 5π/3
D. x =π/3 , 4π/3

2. 10cos x – 5 √(3) = 0

A. x =5π/6 , 7π/6
B. x = π/6 , 11π/6
C. x = 7π/6 , 11π/6
D. x = π/6 , 7π/6

3. 18cos x – 9 = 0

A.  x = 4π/3 , 5π/3
B . x = 2π/3 , 4π/3
C . x = π/3 , 5π/3
D. x = π/3 , 4π/3

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## Exact Values for Trig Functions

If you are in trigonometry or pre-calculus, the below is something you want to memorize, write down, something! Here is a table with the “exact values” for the important angles we use:

 Degrees Radians sin(x) cos(x) tan(x) 0 (or 360) 0 (or 2π) 0 1 0 30 π/6 1/2 √(3)/2 √(3)/3 45 π/4 √(2)/2 √(2)/2 1 60 π/3 √(3)/2 1/2 √(3) 90 π/2 1 0 Undefined 180 π 0 -1 0 270 3π/2 -1 0 Undefined

Remember:  If you ever need to find csc, sec, or cot values, they are just reciprocals.  Csc is the reciprocal of sin, sec is the reciprocal of cos, and cot is the reciprocal of tan.