Trigonometric proof that sum of positive numbers=0

The sum of tangents trigonometric identity gives us that

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β. \tan ( \alpha + \beta) = \frac{ \tan \alpha + \tan \beta} { 1 - \tan \alpha \tan \beta} .

By letting α=tan⁡−1x \alpha = \tan^{-1} x and β=tan⁡−1y \beta = \tan^{-1} y , the equivalent trigonometric identity on tan⁡−1 \tan^{-1} is

tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy). \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y} { 1-xy} \right).

Let's use this identity to calculate tan⁡−11+tan⁡−12+tan⁡−13 \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 . We first have

tan⁡−11+tan⁡−12=tan⁡−1(1+21−1×2)=tan⁡−1(−3). \tan^{-1} 1 + \tan^{-1} 2 = \tan^{-1} \left( \frac { 1 + 2 } { 1 - 1 \times 2} \right)= \tan^{-1} (-3).

As such, this gives

tan⁡−11+tan⁡−12+tan⁡−13=tan⁡−1(−3)+tan⁡−13=tan⁡−1(−3+31−(−3)×3)=tan⁡−10=0 \begin{aligned} & \tan^{-1} 1+ \tan^{-1} 2 + \tan^{-1} 3 \\ = & \tan^{-1} (-3) + \tan^{-1} 3\\ = & \tan^{-1} \left( \frac{-3 + 3} { 1 - (-3)\times 3 } \right) \\ = & \tan^{-1} 0 \\ = & 0 \\ \end{aligned}

By considering the right triangles, we get that tan⁡−11>0 \tan^{-1} 1 > 0 , tan⁡−12>0 \tan^{-1} 2 > 0 and tan⁡−13>0 \tan^{-1} 3 > 0 . Hence, the sum of 3 positive terms is 0.

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#Algebra #Proofs #Olympiad

Note by Calvin Lin
5 years, 3 months ago

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I think you missed the condition there ...... Mod(xy) should be less than one to use the formula .... Or els the RHS becomes negative where as LHS remains positive!!

Abhinav Raichur - 5 years, 2 months ago

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