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Find k

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Find the smallest positive integer $k$ such that

$\tan k^{\circ}=\dfrac{\cos 2020^{\circ}+\sin 2020^{\circ}}{\cos 2020^{\circ}-\sin 2020^{\circ}}$
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
You are going to keep me up all night on this one. Again!

-Dan
 

kaliprasad

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Mar 31, 2013
1,309
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
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
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!:cool:

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
Well done, kaliprasad!
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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
\(\displaystyle \frac{cos(2020) + sin(2020)}{cos(2020) - sin(2020)} = \frac{-cos(40) - sin(40)}{-cos(40) + sin(40)}\)

\(\displaystyle = \frac{cos(40) + sin(40)}{cos(40) - sin(40)}\)

\(\displaystyle = \frac{cos(40) + sin(40)}{cos(40) - sin(40)} \cdot \frac{cos(40) + sin(40)}{cos(40) + sin(40)}\)

\(\displaystyle = \frac{cos^2(40) + 2sin(40)~cos(40) + sin^2(40)}{cos^2(40) - sin^2(40)}\)

After some simplifying:
\(\displaystyle tan(k) = \frac{sin(80) + 1}{cos(80)}\)

Now look at the "averaging formula" for tangent:
\(\displaystyle tan \left ( \frac{\alpha + \beta}{2} \right ) = \frac{sin( \alpha ) + sin(\beta )}{cos( \alpha ) + cos( \beta )}\)

If we let \(\displaystyle \alpha = 80\) and \(\displaystyle \beta = 90\)
\(\displaystyle tan \left ( \frac{80 + 90}{2} \right ) = tan(85) = \frac{sin(80) + 1}{cos(80)} = tan(k)\)

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
 
Last edited:
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
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
\(\displaystyle \frac{cos(2020) + sin(2020)}{cos(2020) - sin(2020)} = \frac{-cos(40) - sin(40)}{-cos(40) + sin(40)}\)

\(\displaystyle = \frac{cos(40) + sin(40)}{cos(40) - sin(40)}\)

\(\displaystyle = \frac{cos(40) + sin(40)}{cos(40) - sin(40)} \cdot \frac{cos(40) + sin(40)}{cos(40) + sin(40)}\)

\(\displaystyle = \frac{cos^2(40) + 2sin(40)~cos(40) + sin^2(40)}{cos^2(40) - sin^2(40)}\)

After some simplifying:
\(\displaystyle tan(k) = \frac{sin(80) + 1}{cos(80)}\)

Now look at the "averaging formula" for tangent:
\(\displaystyle tan \left ( \frac{\alpha + \beta}{2} \right ) = \frac{sin( \alpha ) + sin(\beta )}{cos( \alpha ) + cos( \beta )}\)

If we let \(\displaystyle \alpha = 80\) and \(\displaystyle \beta = 90\)
\(\displaystyle tan \left ( \frac{80 + 90}{2} \right ) = tan(85) = \frac{sin(80) + 1}{cos(80)} = tan(k)\)

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.:eek:

Bravo, Dan!:cool: