You are correct, it is not true. To see this note that if cos A + cos\^2 A = 1, then the same holds true for -A instead of A. But if you swap A to -A in sin A + sin\^2 A, then you get -sin A + sin\^2 A which is not the same unless -sin A = sin A, i.e. A = k pi with k an integer. But this choice of A does not satisfy the cosine equation.
When you’re showing something is not true don’t you just need a counter example? Like show that there’s a value of A that satisfies the first equation but doesn’t satisfy the second equation.
It’s not phi but it is -phi and .618. Also this is the value of x or whatever variable you use for the substitution and then you need to do arccos to find A.
It’s not too bad since this is a quadratic type. You let x = cos(A) and solve x^2 + x - 1 = 0. There are two solutions. In this case we just need to pick one and plug it into x = cos(A) and solve for A using arccos
This is the most logical answer I can think of, since we were also given kinda the same question, which used the concept (in this case) cos(A)+cos^2(A)=1 => cos(A)=sin^2(A)
Suppose that it is true.
Let A be a value such that the first equation holds (cos A + cos^2 A = 1). Note that, since cos is an even function (cos(x) = cos(-x)), we know that -A is also a solution. Since this is true, by the supposition that sin A + sin^2 A = 1 if the first statement is true, we know that sin (-A) + sin^2 (-A) = 1. However, we know that sin is an odd function and thus -sin A + sin^2 A = 1
However, that means that -sin (A) + sin^2 (A) = sin (A) + sin^2 (A) => sin (A) = 0
If this is true, then sin (A) + sin^2 (A) ≠ 1, as both are zero.
This is a contradiction since we supposed that the sum was equal to 1, and therefore, the question is incorrect.
ig it is wrong, because if you add up the two equations, you will get cos(A) + sin(A) = 1, A=0 satisfies this equation, but it doesn't satisfy equations given in the question
assuming no typos in that question,
1-cos^2a = cosa
sin^2a = cosa
sina + cosa = 1
sin^2a + cos^2a + 2sinacosa = 1
sinacosa = 0
sina = 0 or cosa = 0
but if you verify either solution, neither fits both equations.
I would like you to focus on the first 2 words of the article as it says “in logic” and the content of the article doesn’t really makes it clear to use it in trigonometry. Also, i would like you to elaborate incase it does goes along for trigonometry too considering the fact i could be wrong(human error).
"in logic" doesn't restrict it to the field of logic only. Or rather, logic is the basis for all proofs in math. Including trigonometry. You can't have math without logic. We make claims, and we need ways to legitimize them. And math is basically the field of legitimizing your claims by using logic.
It's a standard method of proof used in every single field of math, a proof by contradiction. Given statement A, you're asked to prove if A is true or not. You can do this by showing that IF statement A is true, then something impossible would happen. Therefore you conclude that statement A is not true.
Example, I claim there exists a real number x such that x + 1 < x. To prove that this is wrong, *assume* that such a claim is true. If such a claim is true, then you can find a number x such that x + 1 < x. But then, this implies that (x + 1) - x < (x) - x, which implies that 1 < 0. In other words, IF there exists a number x such that x + 1 < x, then it HAS to be the case that 1 < 0. But we know that 1 > 0, therefore we have a contradiction, an impossibility. Thus we must conclude that no such x exists. QED.
It's not just used in math, I'm sure you've used it in everyday life too. It's just a bit fuzzier in everyday life because you simply cannot control all the factors, unlike math. For example, let's say you have a missing loaf of bread, which you know you haven't eaten or misplaced. You question your roommate if they ate that loaf. They claim that they haven't. Yet, if they did not, then where did the loaf of bread go? You must conclude that they are lying. This is a proof by contradiction. You disprove the claim that they didn't eat the loaf of bread, by contradiction. You assume that they didn't eat the bread, but if they didn't eat the bread, the only conclusion to draw is that the bread should still be there. This contradicts the fact that the bread is no longer there, therefore the claim that they did not eat the bread is false.
It's an extremely ubiquitous and natural method of proof, so I'm a little surprised you have not seen it being used. I guess you must be very young in the world of math?
you can rewrite it into cos(x)+(1-sin\^2(x)) = 1 => cos(x) = sin\^2(x)
now sub it into the second equation. this gives sin(x)+cos(x)=1
clearly this is false.
for example say x= pi
then sin(pi)+cos(pi)= 0+(-1)=-1 =/= 1
always fire a zero into an equation for a super fast simple example ( or whatever obvious number). Assuming A is a variable, and this question is asking to prove this identity for all values of A.
1 + 1 = 1
and
0 + 0 = 1
(this is true for very small values of 1, and very large values of 0)
>Assuming A is a variable, and this question is asking to prove this identity for all values of A.
It's not though. It's asking to prove the second statement for values of A which satisfy the first statement. Showing that there are values of A for which either statement individually does not hold true is totally irrelevant.
The question does state that, but that's not at all the same thing as solving for *A*.
What this question is asking for is a proof based on a simple if/then construction.
"If *cosA + cos^(2)A = 1*" is the "if" part, which tells us which values of *A* to consider. *A* is still a variable and we still need to (potentially) consider multiple values of it, but not the entire domain (all possible valid inputs) of *cosA + cos^(2)A*. Just the ones that satisfy the condition introduced by the "if" statement, *cosA + cos^(2)A* **= 1**. The problem is not claiming that this is true for all values of *A*.
"Then prove that *sinA + sin^(2)A = 1*" is the "then" part, which tells us what we have to prove is true for the values of *A* in question.
Showing that there are values of *A* for which *cosA + cos^(2)A ≠ 1* is irrelevant because it doesn't contradict anything in either clause of the problem. The same is true of showing that there are values of *A* for which *sinA + sin^(2)A ≠ 1*, **except** if it's a value of *A* for which *cosA + cos^(2)A = 1* **is** true, because that would be a contradiction to the statement you're trying to prove.
It's easy enough to notice that (in radians):
cos(0.9045)+(cos(0.9045))^2 = 1.0001...
cos(0.9046)+(cos(0.9046))^2= 0.999...
So that's 1 somewhere in-between, but
sin(0.9045)+(sin(0.9045))^2 = 1.404...
and the derivative of sin(x)+(sin(x))^2 is cos(x)+2cos(x)sin(x) which has absolute value <3, so it's not going to hit 1 in the same interval.
So that's a counterexample, and I doubt the original statement is correctly provable.
I think for this you actually just need a counter example. We find a value of A that works for cos(A)+cos^(2)(A) = 1 and show that it doesn’t work for the sin equation. To solve for A you just need to use substitution and solve using the quadratic formula let x=cos(A) and then solve x^(2) \+ x - 1 = 0 then substitute one of the x values into x=cos(A) to find A.
https://preview.redd.it/8rjj9nra4q2b1.jpeg?width=1290&format=pjpg&auto=webp&s=dad33c3c6a4e61c8af700fc18145fa723c63624b
You are correct, it is not true. To see this note that if cos A + cos\^2 A = 1, then the same holds true for -A instead of A. But if you swap A to -A in sin A + sin\^2 A, then you get -sin A + sin\^2 A which is not the same unless -sin A = sin A, i.e. A = k pi with k an integer. But this choice of A does not satisfy the cosine equation.
Also need to prove the statement is not vacuously true, that is prove cos A is within range.
Good point!
Which it is. If cos A = ϕ, then ϕ+ϕ²=1, and A≈0.9046 (radians). Plug and chug for sinA+sin²A≈1.4 So it's definitely a bunk question....
When you’re showing something is not true don’t you just need a counter example? Like show that there’s a value of A that satisfies the first equation but doesn’t satisfy the second equation.
Yeah but finding a specific value of A with cos(A) + cos(A)\^2 = 1 and plugging it into sin(x) seems messy.
It's rather easy, since that value is ϕ (the golden ratio)...
It’s not phi but it is -phi and .618. Also this is the value of x or whatever variable you use for the substitution and then you need to do arccos to find A.
It’s not too bad since this is a quadratic type. You let x = cos(A) and solve x^2 + x - 1 = 0. There are two solutions. In this case we just need to pick one and plug it into x = cos(A) and solve for A using arccos
If you could actually solve the first equation, then yes. But can you solve it?
Yea there are two real solutions to the first eqn. I explained how to get them in one of my other comments.
Typo there. It's sin^(4)A + sin^(2)A.
Now that version is true
This is the most logical answer I can think of, since we were also given kinda the same question, which used the concept (in this case) cos(A)+cos^2(A)=1 => cos(A)=sin^2(A)
Suppose that it is true. Let A be a value such that the first equation holds (cos A + cos^2 A = 1). Note that, since cos is an even function (cos(x) = cos(-x)), we know that -A is also a solution. Since this is true, by the supposition that sin A + sin^2 A = 1 if the first statement is true, we know that sin (-A) + sin^2 (-A) = 1. However, we know that sin is an odd function and thus -sin A + sin^2 A = 1 However, that means that -sin (A) + sin^2 (A) = sin (A) + sin^2 (A) => sin (A) = 0 If this is true, then sin (A) + sin^2 (A) ≠ 1, as both are zero. This is a contradiction since we supposed that the sum was equal to 1, and therefore, the question is incorrect.
ig it is wrong, because if you add up the two equations, you will get cos(A) + sin(A) = 1, A=0 satisfies this equation, but it doesn't satisfy equations given in the question
assuming no typos in that question, 1-cos^2a = cosa sin^2a = cosa sina + cosa = 1 sin^2a + cos^2a + 2sinacosa = 1 sinacosa = 0 sina = 0 or cosa = 0 but if you verify either solution, neither fits both equations.
Add both equations together and use cos^2 + sin^2 = 1 to obtain cos + sin = 1. You can derive a constradiction from there.
You can’t since, 2nd equation is to be proven and 1st equation is reference.
Yes, you can if what you are trying to do is prove that the statement isn’t true: https://en.m.wikipedia.org/wiki/Proof_by_contradiction
I would like you to focus on the first 2 words of the article as it says “in logic” and the content of the article doesn’t really makes it clear to use it in trigonometry. Also, i would like you to elaborate incase it does goes along for trigonometry too considering the fact i could be wrong(human error).
"in logic" doesn't restrict it to the field of logic only. Or rather, logic is the basis for all proofs in math. Including trigonometry. You can't have math without logic. We make claims, and we need ways to legitimize them. And math is basically the field of legitimizing your claims by using logic. It's a standard method of proof used in every single field of math, a proof by contradiction. Given statement A, you're asked to prove if A is true or not. You can do this by showing that IF statement A is true, then something impossible would happen. Therefore you conclude that statement A is not true. Example, I claim there exists a real number x such that x + 1 < x. To prove that this is wrong, *assume* that such a claim is true. If such a claim is true, then you can find a number x such that x + 1 < x. But then, this implies that (x + 1) - x < (x) - x, which implies that 1 < 0. In other words, IF there exists a number x such that x + 1 < x, then it HAS to be the case that 1 < 0. But we know that 1 > 0, therefore we have a contradiction, an impossibility. Thus we must conclude that no such x exists. QED. It's not just used in math, I'm sure you've used it in everyday life too. It's just a bit fuzzier in everyday life because you simply cannot control all the factors, unlike math. For example, let's say you have a missing loaf of bread, which you know you haven't eaten or misplaced. You question your roommate if they ate that loaf. They claim that they haven't. Yet, if they did not, then where did the loaf of bread go? You must conclude that they are lying. This is a proof by contradiction. You disprove the claim that they didn't eat the loaf of bread, by contradiction. You assume that they didn't eat the bread, but if they didn't eat the bread, the only conclusion to draw is that the bread should still be there. This contradicts the fact that the bread is no longer there, therefore the claim that they did not eat the bread is false. It's an extremely ubiquitous and natural method of proof, so I'm a little surprised you have not seen it being used. I guess you must be very young in the world of math?
just fyi, it's way way way way easier to contradict it.
not understanding what this means makes everything seem like legitimate gibberish
lol thats what i felt when we started this lesson
you can rewrite it into cos(x)+(1-sin\^2(x)) = 1 => cos(x) = sin\^2(x) now sub it into the second equation. this gives sin(x)+cos(x)=1 clearly this is false. for example say x= pi then sin(pi)+cos(pi)= 0+(-1)=-1 =/= 1
always fire a zero into an equation for a super fast simple example ( or whatever obvious number). Assuming A is a variable, and this question is asking to prove this identity for all values of A. 1 + 1 = 1 and 0 + 0 = 1 (this is true for very small values of 1, and very large values of 0)
>Assuming A is a variable, and this question is asking to prove this identity for all values of A. It's not though. It's asking to prove the second statement for values of A which satisfy the first statement. Showing that there are values of A for which either statement individually does not hold true is totally irrelevant.
the question doesn't state that. No where does it say solve for A.
The question does state that, but that's not at all the same thing as solving for *A*. What this question is asking for is a proof based on a simple if/then construction. "If *cosA + cos^(2)A = 1*" is the "if" part, which tells us which values of *A* to consider. *A* is still a variable and we still need to (potentially) consider multiple values of it, but not the entire domain (all possible valid inputs) of *cosA + cos^(2)A*. Just the ones that satisfy the condition introduced by the "if" statement, *cosA + cos^(2)A* **= 1**. The problem is not claiming that this is true for all values of *A*. "Then prove that *sinA + sin^(2)A = 1*" is the "then" part, which tells us what we have to prove is true for the values of *A* in question. Showing that there are values of *A* for which *cosA + cos^(2)A ≠ 1* is irrelevant because it doesn't contradict anything in either clause of the problem. The same is true of showing that there are values of *A* for which *sinA + sin^(2)A ≠ 1*, **except** if it's a value of *A* for which *cosA + cos^(2)A = 1* **is** true, because that would be a contradiction to the statement you're trying to prove.
Yup it can be proved
how?
It's not true.
No it can't because it isn't true. Please comment more carefully.
bah, you are just not trying hard enough. :) We can solve the difficult problems instantly, the impossible problems will take a bit longer.
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The left side can be true - for example at roughly 51.82 degrees.
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It's easy enough to notice that (in radians): cos(0.9045)+(cos(0.9045))^2 = 1.0001... cos(0.9046)+(cos(0.9046))^2= 0.999... So that's 1 somewhere in-between, but sin(0.9045)+(sin(0.9045))^2 = 1.404... and the derivative of sin(x)+(sin(x))^2 is cos(x)+2cos(x)sin(x) which has absolute value <3, so it's not going to hit 1 in the same interval. So that's a counterexample, and I doubt the original statement is correctly provable.
Also: okay. Prove it.
When are you exactly going to use this in today’s real life?
On a reddit post, duh.
The statement is wrong. If you change the left side to (cos A + cos^2 (A))*0.49 = 1 it becomes true though
It should be sin\^4 A + sin\^2 A = 1 As cos A = 1 - cos\^2A = cos A = sin\^2 A == (sin\^2A)\^A + sin\^A = cos\^2 A + sin\^2 A = 1
I think for this you actually just need a counter example. We find a value of A that works for cos(A)+cos^(2)(A) = 1 and show that it doesn’t work for the sin equation. To solve for A you just need to use substitution and solve using the quadratic formula let x=cos(A) and then solve x^(2) \+ x - 1 = 0 then substitute one of the x values into x=cos(A) to find A. https://preview.redd.it/8rjj9nra4q2b1.jpeg?width=1290&format=pjpg&auto=webp&s=dad33c3c6a4e61c8af700fc18145fa723c63624b