The discussion revolves around finding the smallest positive integer k such that tan k° equals a specific trigonometric expression involving cos and sin of 2020°. Dan expresses his satisfaction in solving the problem, acknowledging kaliprasad's earlier solution while noting that it included shortcuts he didn't utilize. Despite the challenge, Dan is eager to share his approach, although he humorously mentions his tendency to take the "hard" way. Other participants show appreciation for Dan's efforts and confirm that their methods are similar. The problem sparked engaging dialogue and camaraderie among the participants.
You are going to keep me up all night on this one. Again!
-Dan
#3
kaliprasad
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We know tan (45 + A) = ( tan 45 + tan A)/( tan 45- tan A)
= ( 1+ sin A/cos A)/( 1- sin A / cos A)
= ( cos A + sin A)/(cos A – sin A)
So ( cos 2020 + sin 2020)/( cos 2020 – sin 2020)
= tan (45 + 2020) = tan (2065) = (tan 2065 mod 180) or tan 85 degrees
Hence k = 85 degrees
#4
anemone
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topsquark said:
You are going to keep me up all night on this one. Again!
-Dan
Hi Dan,
Even though kaliprasad has already cracked it, there are still many other methods to solve the problems and I can't wait to read your solution too!
kaliprasad said:
We know tan (45 + A) = ( tan 45 + tan A)/( tan 45- tan A)
= ( 1+ sin A/cos A)/( 1- sin A / cos A)
= ( cos A + sin A)/(cos A – sin A)
So ( cos 2020 + sin 2020)/( cos 2020 – sin 2020)
= tan (45 + 2020) = tan (2065) = (tan 2065 mod 180) or tan 85 degrees
I finally got it. However kaliprasad's method takes a couple of shortcuts I didn't see so I'm not going to post.
It was a very satisfying problem. Thank you.
-Dan
I'll post it if you wish.
First, an angle of 2020 corresponds to an angle of 220, which is a reference angle of 40 in Quadrant III. So
[math]\frac{cos(2020) + sin(2020)}{cos(2020) - sin(2020)} = \frac{-cos(40) - sin(40)}{-cos(40) + sin(40)}[/math]
After some simplifying:
[math]tan(k) = \frac{sin(80) + 1}{cos(80)}[/math]
Now look at the "averaging formula" for tangent:
[math]tan \left ( \frac{\alpha + \beta}{2} \right ) = \frac{sin( \alpha ) + sin(\beta )}{cos( \alpha ) + cos( \beta )}[/math]
If we let [math]\alpha = 80[/math] and [math]\beta = 90[/math]
[math]tan \left ( \frac{80 + 90}{2} \right ) = tan(85) = \frac{sin(80) + 1}{cos(80)} = tan(k)[/math]
Thus the smallest angle k is thus 85 degrees.
Like I said there are some short-cuts. I tend to do things the "hard" way. (Wink)
-Dan
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#6
anemone
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POTW Director
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topsquark said:
I finally got it. However kaliprasad's method takes a couple of shortcuts I didn't see so I'm not going to post.
It was a very satisfying problem. Thank you.
-Dan
I'll post it if you wish.
First, an angle of 2020 corresponds to an angle of 220, which is a reference angle of 40 in Quadrant III. So
[math]\frac{cos(2020) + sin(2020)}{cos(2020) - sin(2020)} = \frac{-cos(40) - sin(40)}{-cos(40) + sin(40)}[/math]
After some simplifying:
[math]tan(k) = \frac{sin(80) + 1}{cos(80)}[/math]
Now look at the "averaging formula" for tangent:
[math]tan \left ( \frac{\alpha + \beta}{2} \right ) = \frac{sin( \alpha ) + sin(\beta )}{cos( \alpha ) + cos( \beta )}[/math]
If we let [math]\alpha = 80[/math] and [math]\beta = 90[/math]
[math]tan \left ( \frac{80 + 90}{2} \right ) = tan(85) = \frac{sin(80) + 1}{cos(80)} = tan(k)[/math]
Thus the smallest angle k is thus 85 degrees.
Like I said there are some short-cuts. I tend to do things the "hard" way. (Wink)
-Dan
Hi Dan,
Hey, when I said I was looking forward to seeing your solution, I was only joking, as we all know if someone has already cracked a challenge problem, then the chances that others will look into it and solve it differently is very unlikely.
But I appreciate that you solved the problem and posted your solution too and my method is more or less the same as yours.
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