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Find all A*cos(a) + B*cos(b) + C*cos(c) = K where A,B,C,K are integers with GCD 1?
I'd like to find all examples of A*cos(a) + B*cos(b) + C*cos(c) = K
where A, B, C, K are integers whose greatest common divisor is 1,
and where a, b, and c are closed form expressions without inverse trigonometric functions.
Example: 2cos(60) = 2cos(π/3) = 1
But not: 15cos (cos⁻¹(3/5)) + 20cos(cos⁻¹(4/5)) = 25
When their multiplier is non zero, then a, b, and c should be positive and less than a right angle.
Scythian, no restriction on the sign of A, B, C, K. When AB≠0 I can think of an example with C=0 and another with C≠0. Maybe there are more.
Let me revise the question to make it easier: Assume K=0 and a, b, c are non negative and less than right angles.
1 Answer
- 1 decade agoFavorite Answer
Since cos(y)cos(z) = ½(cos(y+z) + cos(y-z)),
multiplying your 2cos(60)=1 by cos(x) yields:
cos(x+60) + cos(60-x) = cos(x)
Also, consider isosceles triangle ABC with apex angle B=b and base angles a. Now place D on AB such that CD=AC=s so that triangle ACD is similar to triangle ABC, angle DCA=b, angle CDA=a, and angle DCB=a-b. Let AC=s.
If BD=CD, then a=2b => 5b=180 or b=36. This means
BC = 2s cos(b) = BA = s + 2s cos(2b) or
2cos(72) + 1 - 2cos(36) = 0
This triangle approach can be extended. Instead of assuming BD=CD to get have two isosceles triangles within ABC, assume point E on BC such that ED=EB=s, forming three isosceles triangles. Then angle EDB = b => 2b = angle CED = angle ECD = a-b, from which 7b=180. Now from AB=BC one may read off:
1 + 2cos(2b) = 2cos(b) + 2cos(3b)
which does not meet the conditions of the revised problem.
Finally, placing 4 isosceles triangles with one more point F so that s=AC=CD=DE=EF=FB, summing angle ADC=a, angle BDE=2b, and angle CDE=180+2b-2a to give 180 yields 4b=a or b=20. Reading off from AB=BC:
2cos(b) + 2cos(3b) = 1 + 2cos(2b) + 2cos(4b) and since 2cos(3b)=1, this simplifies to
cos(20) = cos(40) + cos(80)
which also follows from plugging in x=20 above.
So the first two results above give all the unique solutions I could find. Another way to express the first result is to let x=a+b where a+b=60. This yields:
cos(2a) + cos(2b) = cos(a-b), where a+b=60
