It should be squared
True or False :
( cos x − sin x ) ( cos x + sin x ) = cos 4 x − sin 4 x
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4 solutions
Yes, I was able to understand the solution but it is only applicable in a case where we begin by simplifying RHS and then equating it to LHS. It doesn't work the other way round without adding 1 on both sides, why is that so?
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No, no, you're not adding 1, you're multiplying by 1.
( cos x − sin x ) ( cos x + sin x ) = cos 4 x − sin 4 x
( cos x − sin x ) ( cos x + sin x )
cos 2 x − sin 2 x
( cos 2 x − sin 2 x ) ⋅ 1
( cos 2 x − sin 2 x ) ( cos 2 x + sin 2 x )
cos 4 x − sin 4 x
I was thinking the other way around: (cos x-sin x)(cos x+sin x) = cos 2x. But then that wouldn't be that helpful, so you would reread the problem and see Jack's solution.
The problem is to be solved akhil Bansal's way but another thing which can be done is inputing known values of sin and cos like 0 30 45 60 and 90 and checking if LHS=RHS (Useful for objective type questions)
That may work for some questions, but its not really a good habbit to do so.
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Actually it's very good to plug in these values . But the condition is you should take defined values . Let's say if I put 45 degree as theta , then its always defined under both sin and cos , hence its a genuine and less time consuming way to solve like this .
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This is not a "solving" problem. The goal is to prove or disprove the statement. If you plug in five different values of x and the statement is true for these five values of x, you have only shown that the statement is sometimes true. To show that the statement is always true, you need to use valid mathematical manipulations until you have an identity. The "plug in some test cases" strategy is only useful to show that something is not true. If you can find one test case that results in a contradiction, then you have shown that the statement is not true.
For example, some may think that the square root of x^2 simplifies to x. I can find thousands of test cases for which this is true (any non-negative number). But having thousands of individually correct test cases does not allow you to conclude that that statement is true for any x (try a negative).
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@Kyle Misiaszek – Dude I am and 18 yr old guy who understands a bit of maths . First of all go and study functions again. √x^2 would always open as |x|. Moreover i am not dumb that I won't check the domain , values etc .
What do you think that √4 is always 2 ??? You have to define first the domain is what kind of nos are you talking about
In maths there is something called mathematical induction where you prove for a certain value and then generalise . But acc to you that thing doesn't work and is wrong . Next time do your homework well
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@Sanjwal Singhs – My first comment was not meant to be an insult, so relax. It was simply meant to constructively correct your incorrect statement that plugging in values is " a genuine and less time consuming way to solve like this."
"But acc to you [mathematical induction] doesn't work and is wrong." I am familiar with mathematical induction and I think it's a wonderful way to prove many statements. The statement being discussed can't be proven using induction (since x has a domain of all real numbers, not the whole numbers), which is why I didn't bring up induction. What you suggested, plugging in a few values, is not induction. Proving a base case and proving an induction step is induction.
In summary, plugging in a few values is not a way to prove that a statement is always true. And this Brilliant problem is asking us to prove or disprove a statement, not to (as you put it) solve.
@Sanjwal Singhs – It seems (from my point of view) sort of like creating a proof. You want to validate or invalidate the statement. But your way of testing those 3 numbers only proves that those three work. While you may be able to infer that other numbers may work as well, it doesn't explain the statement and why it works or even if it works for all numbers for x.
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@Anmol Maini – INORDER to prove me wrong you have to find a func which is non continuous on the values ie non continuous on 0 30 45 60 90 . by definition of sin and cos are well defined in context of the question
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@Sanjwal Singhs – But you never proved it right. You showed examples where it worked but never fully proved it to a point that could validate its whole statement. So you are asking me to disprove something that was never fully proved. You know it works but didn't show it and only used examples as evidence for its success.
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@Anmol Maini – Exactly. If you use several values for which the original statement is true, you may suspect that the original statement is true for all values, but suspecting something is not proving something.
That being said, there is a time and a place for Sanjwal's strategy. If this were a question on a test, it might be a good idea to go with your suspicions for the sake of time. But if you are a teacher making an answer key to this problem (or just posting the justification for the answer of TRUE on this site), you need a valid proof not several test cases.
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@Kyle Misiaszek – Kyle is exactly right. I have a Ph.D. in math from the University of Texas and I've been a math professor for 44 years. That does not mean I'm right! But it may mean I have some credibility. To subscribe to my puzzle-solvers list, send e-mail to dgittinger@alamo.edu
@Kyle Misiaszek – Okay I get it a bit but this strategy is real good and saves a minute or two in an exam . Moreover I ALREADY have mentioned that the correct way is the way Akhil did it plz read the first line of my solution
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@Sanjwal Singhs – Depending on the values that you choose to test, you may inadvertently choose values where different functions happen to have the same output. This is what the other responders were trying to indicate. If you take sinx and 2sinx you get matching values if you choose certain texts even though the functions are not equivalent.
Proof in mathematics is rigorous. You cannot prove that functions are equivalent over their entire domain by testing even an infinitely large proper subset of their domain.
@Sanjwal Singhs – I understand that this would work on a test, but this question isn't like that of a test question. It asks whether the statement is true or not, where you would need an existing proof as Luther talks about. You could disprove the proof through examples, but not prove it since as Luther said, you would need to test an infinite domain if you wanted to prove it through examples and even then without concrete proof, you still cannot prove the statement to be true since there would still be more to test since infinity goes on forever
That would be good for harder problems where the result is not so obvious to see. Besides, Jack's method is a lot easier than plugging in nontrivial values of x (like 30).
Correct, that's is what i usually prefer in my objective tests.
Doesn't "foiling" it work all the same?
While it is valid to distribute the two binomials on the left hand side, at some point you will need to factor the right hand side and use the Pythagorean Identity as Akhil did. "FOILing" alone is not enough.
cos 4 x − sin 4 x = ( cos 2 x + sin 2 x ) ( cos 2 x − sin 2 x ) [ cos 2 x + sin 2 x = 1 ]
The expression in the first set of parentheses is a Pythagorean identity that equals 1, leaving us with cos 2 x − sin 2 x = ( cos x + sin x ) ( cos x − sin x )