#### MethMath11

##### Junior Member

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- Thread starter MethMath11
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You can then solve to find all solutions for A and B. (All this can be practically "by inspection" if you think about it the right way, but it can be more formal. There are some subtleties you might miss, but they turn out not to cause trouble.)

I don't see a way to solve this without finding A and B.

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Give that a go.

\(\displaystyle w = sin(a),\ x = sin(b),\ y = cos(a), \text { and } z =cos(b).\)

Then from the given equations, we can work with:

\(\displaystyle w + x = 1,\ y + z = 0, \ w^2 + y^2 = 1, \text { and } x^2 + z^2 = 1.\)

\(\displaystyle w + x = 1 \implies w = 1 - x \implies w^2 = 1 - 2x + x^2.\)

\(\displaystyle y + z = 0 \implies z = -\ y \implies z^2 = y^2.\)

\(\displaystyle w^2 + y^2 = 1 \implies 1 - 2x + x^2 + y^2 = 1 \implies x = \dfrac{2 \pm \sqrt{4 - 4y^2}}{2} = 1 \pm \sqrt{1 - y^2}. \)

\(\displaystyle x^2 + z^2 = 1 \implies x^2 + y^2 = 1 \implies x = \pm \sqrt{1 - y^2}.\)

\(\displaystyle \therefore \pm \sqrt{1 - y^2} = 1 \pm \sqrt{1 - y^2}.\)

If the signs are the same, this is impossible. So the signs must be different.

\(\displaystyle \pm 2 \sqrt{1 - y^2} = 1 \implies 4(1 - y^2) = 1 \implies y^2 = \dfrac{3}{4} \implies y = \pm \dfrac{\sqrt{3}}{2}.\)

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MarkFL suggested trying to spot A and B, so naturally I took the given equations (evaluated) to be 1/2+1/2=1 and -1+1=0, respectively, because I'm familiar with those special-angle trig values. The unit-circle concept memorized helps reveal A and B readily.

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Yes, I looked for two angles symmetric about \(\displaystyle \frac{\pi}{2}\), and readily found 2 angles satisfying both equations.

MarkFL suggested trying to spot A and B, so naturally I took the given equations (evaluated) to be 1/2+1/2=1 and -1+1=0, respectively, because I'm familiar with those special-angle trig values. The unit-circle concept memorized helps reveal A and B readily.

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Mark FL yes I did that too and you get the answer quite easily. I just wasn't 100% sure if that was a complete solution. What if there were other values of A and B?Yes, I looked for two angles symmetric about \(\displaystyle \frac{\pi}{2}\), and readily found 2 angles satisfying both equations.

I suppose the nature of the question "If this … then this" implies that even if you only find one solution to the original 2 simultaneous equations (and ignore any others) then the next part is true. Of course that is relying on the last statement being true for all possible solutions of A and B, and that is relying on author of the question.

I hope that makes sense?? Here' an example of what I'm getting at:

If A

Isn't your logic used above the same as "Oh I can see two answers that work namely A=2 and B=3 so then A+B=5" missing something? Obviously the "then" statement is not correct for all A and B.

I think I agree with myself!!! Thoughts?

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I too saw the obvious answers but were there others? I still think that this problem was a difficult one although JeffM made it look easy.Mark FL yes I did that too and you get the answer quite easily. I just wasn't 100% sure if that was a complete solution. What if there were other values of A and B?

I suppose the nature of the question "If this … then this" implies that even if you only find one solution to the original 2 simultaneous equations (and ignore any others) then the next part is true. Of course that is relying on the last statement being true for all possible solutions of A and B, and that is relying on author of the question.

I hope that makes sense?? Here' an example of what I'm getting at:

If A^{2}= 4 and B^{2}= 9 then A + B = ?.

Isn't your logic used above the same as "Oh I can see two answers that work namely A=2 and B=3 so then A+B=5" missing something? Obviously the "then" statement is not correct for all A and B.

I think I agree with myself!!! Thoughts?

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Yes there will be others because of the periodic nature of the functions involved.I too saw the obvious answers but were there others? I still think that this problem was a difficult one although JeffM made it look easy.

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It's multiple choice, so we only need one.… saw the obvious answers but were there others? …

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Yeah I know that but "maybe" another value is also valid. (It's not in this case but could be.)

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Yes, as long as you only want to get the "right" answer, and trust that they got it correct.It's multiple choice, so we only need one.

If you care about truth, instead, you have to consider that maybe there is no answer, because the value is not determined by the given facts. And if there were a choice like "insufficient information", you would have to dig deeper.

This is one of the reasons I hate multiple choice! It distorts your perspective on math, removing incentive to look for certainty.

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Initially, I looked at squaring the given equations and trying to derive an answer based solely on that. But, in the end, I found I agree with this:Mark FL yes I did that too and you get the answer quite easily. I just wasn't 100% sure if that was a complete solution. What if there were other values of A and B?

I suppose the nature of the question "If this … then this" implies that even if you only find one solution to the original 2 simultaneous equations (and ignore any others) then the next part is true. Of course that is relying on the last statement being true for all possible solutions of A and B, and that is relying on author of the question.

I hope that makes sense?? Here' an example of what I'm getting at:

If A^{2}= 4 and B^{2}= 9 then A + B = ?.

Isn't your logic used above the same as "Oh I can see two answers that work namely A=2 and B=3 so then A+B=5" missing something? Obviously the "then" statement is not correct for all A and B.

I think I agree with myself!!! Thoughts?

I would have much preferred to be able to algebraically derive the double-angle expression from the two given equations, but I was unable to, given the amount of time I could afford to spend on it.I don't see a way to solve this without finding A and B.

The problem says "suppose A and B are two angles such that..." and imply that only one of the given choices can be correct. Having found two such angles, and that one of the answers is the result, I was satisfied.

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I do understand tutors' interest in the side discussion; however, my desire is for MethMath to understand what's… Yes, as long as you only want to get the "right" answer …

If you care about truth, instead, …

I started off as Mark did, but when that did not work, I did not next look for A and B.Initially, I looked at squaring the given equations and trying to derive an answer based solely on that. But, in the end, I found I agree with this:

I would have much preferred to be able to algebraically derive the double-angle expression from the two given equations, but I was unable to, given the amount of time I could afford to spend on it.

The problem says "suppose A and B are two angles such that..." and imply that only one of the given choices can be correct. Having found two such angles, and that one of the answers is the result, I was satisfied.

I fully agree with otis that experimenting with special right triangles as a speedy first approach to apparently difficult trigonometric problems is undoubtedly what the person writing the exercise intended to teach. I further agree with otis that someone who is studying for an exam in trigonometry will often profit from that approach. It did not occur to me to start there because I have not had to take an exam in trigonometry for well over 50 years.

So yes, the lesson for the student is: if a problem in a text or examination on trigonometry seems weird, see whether a special right triangle gives the answer. That is a great lesson for taking exams. I do not, however, see it teaching much in the way of mathematical reasoning.

What I saw was four unknowns and two equations. That lead me to ask myself whether I could find two additional equations. That in turn led me to one of the basic trigonometric identities, namely

\(\displaystyle sin^2( \theta ) + cos^2( \theta ) = 1.\)

That approach, can we find apparently missing equations and can we take advantage of basic trig identities, strikes me as being a more useful general approach for learning mathematics. I see now that I skipped over both points in my answer so the student could not see those ideas in my initial response.

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