Trigonometric proof that sum of positive numbers=0

The sum of tangents trigonometric identity gives us that

tan(α+β)=tanα+tanβ1tanαtanβ. \tan ( \alpha + \beta) = \frac{ \tan \alpha + \tan \beta} { 1 - \tan \alpha \tan \beta} .

By letting α=tan1x \alpha = \tan^{-1} x and β=tan1y \beta = \tan^{-1} y , the equivalent trigonometric identity on tan1 \tan^{-1} is

tan1x+tan1y=tan1(x+y1xy). \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y} { 1-xy} \right).

Let's use this identity to calculate tan11+tan12+tan13 \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 . We first have

tan11+tan12=tan1(1+211×2)=tan1(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

tan11+tan12+tan13=tan1(3)+tan13=tan1(3+31(3)×3)=tan10=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 tan11>0 \tan^{-1} 1 > 0 , tan12>0 \tan^{-1} 2 > 0 and tan13>0 \tan^{-1} 3 > 0 . Hence, the sum of 3 positive terms is 0.

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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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