Question

help Me!
(cos Θ − cos Θ)2 + (cos Θ + cos Θ)2

sin2 Θ

4cos2 Θ

8

cos2 Θ

Answer 1

(cos Θ − cos Θ)^2 + (cos Θ + cos Θ)^2

Answer 2

Hello! Please, share the instructions that came with this problem. Also, plese explain what you mean by those '2's. (OK: I see you've realized tht you must use '^' to denote exponentiation.)

Answer 3

What is your goal in this problem? what have you thought of doing first?

Answer 4

thats the question and choices

Answer 5

simplfy both (cos Θ − cos Θ)^2

Answer 6

surely there are instructions that came with the question.

Answer 7

Yes. You want to 'simplify' the given expression.
What have you considered doing first?

Answer 8

turning it into cos^2 theta -cos^2 theta to both side

Answer 9

would it equal sin^2 theta + sin^2 theta

Answer 10

Your (cos Θ − cos Θ)^2 + (cos Θ + cos Θ)^2 doesn't have two sides; what you're trying to say is that this expression HAS TWO TERMS.

Let me answer your question with a question: What is (a-b)^2? Think before you answer this.
What is (a+b)^2?

Answer 11

a^2-ab-ba+b^2

Answer 12

a^2+ab+ba+b^2

Answer 13

@xapproachesinfinity is the the answer d

Answer 14

It is b, what (a-a)^2 and (a+a)^2?

Answer 15

Please simplify:

a^2-ab-ba+b^2

a^2+ab+ba+b^2

these are called "special products" and you should know these formulas.

Once you've simplified and have both formulas correct, apply them to

(cos Θ − cos Θ)^2 + (cos Θ + cos Θ)^2

Answer 16

how did you get that?

Answer 17

Please do not deal in answers alone. In the end it is up to @helpmeprettyplease4 to find his/her own answers, with our guidance.

Answer 18

(cos Θ − cos Θ)^2 is the same as saying (a-a)^2

Answer 19

Also, "what (a-a)^2 and (a+a)^2?" is an inappropriate question, mathematically incorrect.l

Answer 20

Regarding "(cos Θ − cos Θ)^2 is the same as saying (a-a)^2": I disagree.

Answer 21

@HelpMePrettyPlease4 :

Please simplify:

a^2-ab-ba+b^2

a^2+ab+ba+b^2

these are called "special products" and you should know these formulas.

Once you've simplified and have both formulas correct, apply them to

(cos Θ − cos Θ)^2 + (cos Θ + cos Θ)^2

Answer 22

@mathmale my bad! and I agree that the step i did is incorrect to approach this. I just realized it! thanks for the correction. I usually don't give answers actually

Answer 23

@xapproachesinfinity : Thank you for this mature and honest reply.

Since @HelpMePrettyPlease4 has asked for your help, would you mind guiding him/her towards solving his/her original problem?

Answer 24

a^2-ab+b^2
a^2+ab+b^2

Answer 25

@xapproachesinfinity : Your input?

Answer 26

it is my pleasure! thanks again

Answer 27

As explained earlier to @HelpMePrettyPlease4 : I need to get off OpenStudy. I will be available again sometime later today, if needed.

Answer 28

ok thanks for your help @mathmale :-)

Answer 29

I do not understand why any binomial is being considered.

cos(x) - cos(x) = 0 -- Square it? Still zero.

cos(x) + cos(x) = 2cos(x) -- Square it? 4cos^2(x)

Done.

Answer 30

No that is a incorrect as he said! I agree

Answer 31

ohhh wow thats looks way easier!

Answer 32

that was my initial mistake that i used to approach this

Answer 33

wait @tkhunny is wrong?

Answer 34

No it easy but incorrect way of doing it. i have done the same thing to get B, and i told you the result right?

Answer 35

yeah, so what is the correct way to do it?

Answer 36

No, tkhunny is not wrong. There is no "right" or "correct" way to do it. Try to simplify your life, when you can. Unique answers don't care how you find them.

Answer 37

so let's go back to the original thing
he said that (a-b)^2= a^2-ab-ba+b^2
and (a+b)^2=a^2+ab+ba+b^2
then we simplify
(a-b)^2=a^2-2ab+b^2
(a+b)^2=a^2+2ab+b^2
so we use this the solve that

Answer 38

@tkhunny so the answer is 4cos^2 Θ

Answer 39

Like I said, you can do it that way if you like, there is just 500% more to go wrong. Be very, very careful and you can find the right answer that is much more simply generated by a few simple observations.

Stop asking for the answer. PROVE it, mark it, and move on. You tell me what the right answer is.

Answer 40

@tkhunny , I have been told it is incorrect! let's discuss that on a post later!
with everyone involved

Answer 41

I'm not so sure about it! haha

Answer 42

ty

Answer 43

Awesome. I'm glad you listened when you were told it was wrong. You should have argued. The fact is, cos(x) - cos(x) = 0. There is no adverse consequence to the substitution.

Pick an angle:
\((\cos(\theta) - cos(\theta))^{2} = 0^{2} = 0\)

There is NO counterexample to this conclusion.

Answer 44

you will get cos^2-2cos+cos^2 +cos^2+2cos^2+cos^2
-2cos^2 + 2cos^2 cancels we are left with cos^2+cos^2+cos^2+cos^2

Answer 45

Again, if you want to multiply it out, go right ahead. If you are careful, you will reach the same, unique result.

Answer 46

ty so much! :-) i appreciate all of you guys help!

Answer 47

well I heard the argument, but I don't remember what it was haha.

Answer 48

I agree with tkhunny.
cos(x) - cos(x) = 0 for all x.

But the converse is not always true:
cos(A) - cos(B) = 0 does not necessarily imply A = B.

Answer 49

And it's not really an argument. No one is talking about A = B in some general sense or any implication about A and B not equal. The problem statement assumes only A = A. There is, therefore, no additional discussion to be had.

Answer 50

You closed it all! @tkhunny, thanks! it is good to have this kind of discussion.
I will re check about this and see. I have seen something similar to this. I could be wrong

Answer 51

Not everything has to be an argument. The first paragraph of my previous reply applies to this problem. The second paragraph is just an FYI for folks because I have seen the mistaken conclusion made by some.