Question

Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos(x + pi/4 ) − cos(x - pi/4 ) =1

Answer 1

Use the formulas:
(1) \(\color{blue}{\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) }\)
(2) \(\color{blue}{\displaystyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) }\)

Answer 2

(Use them to rewrite your equation in terms of \(\sin(x)\) and \(\cos(x)\).)

Answer 3

Based on the rules above, can you re-write \(\color{black}{\displaystyle \cos(x+\pi/4)}\) ?

Answer 4

Yes!

Answer 5

Yes, you can? Good! Can you please show me how you would do this?

Answer 6

cos(a−b)=cos(a)cos(b)+sin(a)sin(b)

cos(x+π/4)=cos(x)cos(π/4) +sin(x)sin(π/4)

Answer 7

Check your signs please. (You used the wrong property, it's a\(\color{red}{+}\)b.)

Answer 8

We know that: \(\color{blue}{\displaystyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b) }\)

So, \(\color{black}{\displaystyle \cos(x+\pi/4)=? }\)

Answer 9

cos(x+π/4)=cos(x)cos(π/4) -sin(x)sin(π/4)

Answer 10

yes, fabulous!

Answer 11

Can you also rewrite \(\color{black}{\displaystyle \cos(x-\pi/4) }\) for me please?

Answer 12

cos(x-π/4)=cos(x)cos(π/4) +sin(x)sin(π/4)

Answer 13

Yes, excellent!

Answer 14

So, using the properties above we have:
1. \(\color{black}{\displaystyle \cos(x+\pi/4)=\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4)}\)
2. \(\color{black}{\displaystyle \cos(x-\pi/4)=\cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4)}\)

Answer 15

Now, go ahead and substitute the above results into the original
equation (i.e. into \(\color{black}{\displaystyle \cos(x+\pi/4)-\cos(x-\pi/4)=1}\)).

Answer 16

\(\large[\)Basically, replace \(\color{black}{\displaystyle \cos(x+\pi/4)}\) with \(\color{black}{\displaystyle \cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4)}\),
and replace \(\color{black}{\displaystyle \cos(x-\pi/4)}\) with \(\color{black}{\displaystyle \cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4)}\).\(\large]\)

Answer 17

(If you need help substituting, let me know)

Answer 18

:( I want to cry because my head is about to burst

Answer 19

\(\LARGE \color{black}{\displaystyle \color{black}{\displaystyle \color{red}{\underbrace{\color{black}{\cos(x+\pi/4)}}_{\large \cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4)}}-\color{red}{\underbrace{\color{black}{\cos(x-\pi/4)}}_{\large \cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4)}}=1}}\)

Answer 20

Ok, I will sub for you:)

Answer 21

So you just plugged it in?

Answer 22

\(\color{black}{\displaystyle \cos(x+\pi/4)-\cos(x-\pi/4)=1}\)

\(\color{black}{\displaystyle {\large [}\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4){\large ]}-{\large [}\cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4){\large ]}=1}\)

Answer 23

Yes, all we need is just to plug it in.
(Substitute in the same thing, basically)

Answer 24

Well, in this context, plug and sub is exactly the same.

Answer 25

Anyway, do you see why we get
\(\color{black}{\displaystyle {\large [}\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4){\large ]}-{\large [}\cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4){\large ]}=1}\)

??

Answer 26

Yes!

Answer 27

Ok, can you distribute the negative in front of the 2nd bracket, and simplify?

Answer 28

(We know that \(\color{black}{\displaystyle -(a+b)=-a-b}\), right?)

Answer 29

[-cos(x)cos(π/4)-sin(x)sin(π/4)]=1 right?

Answer 30

yes!

Answer 31

Yes!

Answer 32

So, now instead of
\(\color{black}{\displaystyle {\large [}\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4){\large ]}-{\large [}\cos(x)\cos(\pi/4)+\sin(x)\sin(\pi/4){\large ]}=1}\)

we have
\(\color{black}{\displaystyle {\large [}\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4){\large ]}-\cos(x)\cos(\pi/4)\color{blue}{-} \sin(x)\sin(\pi/4)=1}\)

Answer 33

which is same as
\(\color{black}{\displaystyle \cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4)-\cos(x)\cos(\pi/4)-\sin(x)\sin(\pi/4)=1}\)

(without the first brackets)

Answer 34

Do you see anything that cancels out here?

Answer 35

The sine right?

Answer 36

The sines are both negative. They don't cancel.

Answer 37

Wait! The cosine!!!! Sorry I need to look more carefully!

Answer 38

yes, so what is your new equation?

Answer 39

−sin(x)sin(π/4)−sin(x)sin(π/4)=1

Answer 40

yes, \(\color{black}{\displaystyle -\sin(x)\sin(\pi/4)-\sin(x)\sin(\pi/4)=1}\)

Answer 41

Well, as you know \(\color{black}{\displaystyle -b-b=-2b}\), correct?

Answer 42

Similarly, you have the like-terms in this situation ... how can you simplify your equation?

Answer 43

Correct!

Answer 44

−2sin(x)sin(π/4)=1

Answer 45

Yes, right!!

Answer 46

Do you know what \(\color{black}{\displaystyle\sin(\pi/4)}\) is equal to?

Answer 47

I don't. I'm sorry! :(

Answer 48

\(\color{black}{\displaystyle \sin(\pi/4)=\sin(45^\circ)}\)

Answer 49

How did you get that?

Answer 50

Unit circle?

Answer 51

That is the same thing.
This is conversion from radians to degrees.

Answer 52

Oh!!! I know how to do that

Answer 53

Yes, good;)

Answer 54

So, can you tell me what \(\color{black}{\displaystyle \sin(\pi/4)}\) (or sin(45\(^\circ\)) is equal to?

Answer 55

If not, let me know, and we can deduce it.

Answer 56

\[\frac{ \sqrt{2} }{ 2 }\]

Answer 57

Yup, correct!

Answer 58

We have \(\color{black}{\displaystyle -2\sin(x)\sin(\pi/4)=1}\)

Answer 59

Can you re-write this, knowing that \(\color{black}{\displaystyle \sin(\pi/4)=\frac{\sqrt{2}}{2}}\) ?

Answer 60

Substitute it in what?

Answer 61

\(\color{black}{\displaystyle -2{\tiny~}\sin(x){\tiny~}\color{red}{\underbrace{\color{black}{\sin(\pi/4)}}_{\Large \frac{\sqrt{2}}{2}}}=1}\)

Answer 62

You are substituting \(\color{black}{\displaystyle \frac{\sqrt{2}}{2}}\) instead of \(\color{black}{\displaystyle\sin(\pi/4)}\), knowing that \(\color{black}{\displaystyle\sin(\pi/4)=\sqrt{2}/2}\).

Answer 63

Oh!!! Okay that makes sense!

Answer 64

Awesome!
Now, please go ahead and substitute (plug) this.

Answer 65

−2 sin(x) ( √2/2 ) =1

Answer 66

Yes, correct !

Answer 67

\(\color{black}{\displaystyle -2\sin(x)\cdot \frac{\sqrt{2}}{2}=1}\)

Answer 68

Can you simplify this one step further?

Answer 69

I do not see anything that cancels out

Answer 70

\(\color{black}{\displaystyle -\color{red}{2}\sin(x)\cdot \frac{\sqrt{2}}{\color{red}{2}}=1}\)

Answer 71

How about now? :)

Answer 72

The 2's

Answer 73

Yup, so what equation do you get now?

Answer 74

sin(x) squaroot 2 =1

Answer 75

You forgot the negative in the beginning.

Answer 76

-sin(x) squaroot 2 =1

Answer 77

Yup!

Answer 78

\(\color{black}{\displaystyle -\sqrt{2{\tiny~}}{\tiny~}\sin(x)=1}\)

Answer 79

So what would the final answer be?

Answer 80

There is just a little work left to figure this out.

Answer 81

Can you isolate \(\sin(x)\) for me please?

Answer 82

(What would you divide by, on both sides, in order to get \(\sin(x)\) on the right side alone?)

Answer 83

\[\frac{ 1 }{ \sqrt{2} }\]

Answer 84

negative

Answer 85

When simplified?

Answer 86

Are you saying \(\color{black}{\displaystyle {\tiny~}\sin(x)=-\frac{1}{\sqrt{2{\tiny~}}}}\) ?

Answer 87

yes

Answer 88

Yes, that is right once again!

Answer 89

When you rationalize the denominator,

\(\color{black}{\displaystyle {\tiny~}\sin(x)=-\frac{1}{\sqrt{2{\tiny~}}}}\)

\(\color{black}{\displaystyle {\tiny~}\sin(x)=-\frac{1\times \sqrt{2{\tiny~}}}{\sqrt{2{\tiny~}}\times \sqrt{2{\tiny~}}}}\)

\(\color{black}{\displaystyle {\tiny~}\sin(x)=-\frac{\sqrt{2}}{2}}\)

Answer 90

K?

Answer 91

Yes!

Answer 92

If it were to be \(\color{black}{\displaystyle {\tiny~}\sin(x)=\frac{\sqrt{2}}{2}}\),
then you would get \(\color{black}{\displaystyle x=...~-7\pi/4~~\pi/4,~~9\pi/4~~...}\)

Answer 93

So that is the final answer?

Answer 94

Read what I say carefully, please. I said if sin(x) .... that is not an equation, I am mentioning that case because it relates to the solution.

Answer 95

It is an equation (in general), but it is not OUR equation.
(It relates to answer, this is why I mentioned it.)

Answer 96

Oh excuse me I got it incorrectly. I made a mistake here just now.

Answer 97

5pi/4 and 7pi/4

Answer 98

and yes you are correct, good job.