I recently determined that for all integers a and b such that a≠b and b≠0.arctan⁡(ab)+π4=arctan⁡(b+ab−a)

Question

Abbey Hope · Accepted Answer

As written, the formula is not true: the values of arctan⁡(x) are always between −π2 and π2. Pick a rational number ab with π4&amp;lt;ab&amp;lt;π2. For example, a=11, b=10. Then the left hand side,arctan⁡(1110)+π4≈1.6184whereas the right hand side is negative:arctan⁡(11+1010−11=arctan⁡(−21)≈−1.5232I think that what you mean is that if α is an angle such that tan⁡(α) is rational, different from 1,tan⁡(α)=ab≠1, a,b integers,thentan⁡(α+π4)=b+ab−aCertainly, well done if you discovered it by yourself! However, it is not new. In fact, the result is true even if a and b are not integers; all you need is for a to be different from b, that is, for α≠π4There are well-known formulas that express the sine, cosine, and tangent of a sum of angles in terms of the sines, cosines, and tangents of the summands:sin⁡(α+β)=sin⁡(α)cos⁡(β)+cos⁡(α)sin⁡(β)cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)tan⁡(α+β)=tan⁡(α)+tan⁡(β)1−tan⁡(α)tan⁡(β)Taking β=π4, since tan⁡(π4)=1, we gettan⁡(α+π4)=ab+11−ab=a+bbb−ab=a+bb−agiving your formula.

e4mot1ic5bf · Accepted Answer

If you differentiate the functionf(t)=arctan⁡t−arctan⁡1+t1−tyou get zero, so the function is constant in each of the two intervals (−∞,1) and (1,+∞) on which it is defined.Its calue at zero is π2, so that f(t)=−π4 for all t&amp;lt;1, soIts value at zero is π2, so that f(t)=−π4 for all t&amp;lt;1, soarctan⁡t+π4=arctan⁡1+t1−t,On the other hand, one easily shows that limt→+∞f(t)=3π4, soarctan⁡t−3π4=arctan⁡1+t1−tIf t=ab is a rational number smaller that 1, then the first point is your identity. If it larger than 1, we see that you have to change things a bit.

I recently determined that for all integers a and b

Answered question

Answer & Explanation

New Questions in Precalculus