Sum of inverse tangents of roots

Geometry Level 4

If x 1 , x 2 , x 3 , x 4 x_1, x_2, x_3, x_4 are rotos to the equation x 4 − ( sin ⁡ 2 θ ) x 3 + ( cos ⁡ 2 θ ) x 2 − x cos ⁡ θ − sin ⁡ θ = 0 x^4 - (\sin2\theta) x^3 + (\cos 2\theta) x^2- x \cos\theta - \sin\theta = 0 , compute ∑ i = 1 4 tan ⁡ − 1 x i \displaystyle \sum_{i=1}^4 \tan^{-1} x_i .

− θ -\theta 9 0 o − θ 90^{o} - \theta θ \theta 4 5 o − θ 45^{o} - \theta

Sanghamitra Anand by Sanghamitra Anand , 21, India

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1 solution

We know ∑ i = 1 4 t a n − 1 x i = t a n − 1 S 1 − S 3 1 − S 2 + S 4 \sum_{i=1}^{4}tan^{-1}x_i = tan^{-1}\frac{S_1-S_3}{1-S_2+S_4}

Where S k S_k denotes the sum of the variables taken k at a time.

Here S 1 = s i n ( 2 θ ) , S 2 = c o s ( 2 θ ) , S 3 = c o s ( θ ) , S 4 = − s i n ( θ ) S_1 = sin(2\theta), S_2 = cos(2\theta) , S_3 = cos(\theta) , S_4 = -sin(\theta) [As we know from vieta's]

So , ∑ i = 1 4 t a n − 1 x i = t a n − 1 s i n ( 2 θ ) − c o s ( θ ) 1 − c o s ( 2 θ ) − s i n ( θ ) \sum_{i=1}^{4}tan^{-1}x_i = tan^{-1} \frac{sin(2\theta)-cos(\theta)}{1-cos(2\theta)-sin(\theta)}

= t a n − 1 c o s ( θ ) ( 2 s i n ( θ ) − 1 ) 2 s i n 2 ( θ ) − s i n ( θ ) = tan^{-1} \frac{cos(\theta)(2sin(\theta)-1)}{2sin^2(\theta)-sin(\theta)}

= t a n − 1 c o s ( θ ) ( 2 s i n ( θ ) − 1 ) s i n ( θ ) ( 2 s i n ( θ ) − 1 ) =tan^{-1} \frac{cos(\theta)(2sin(\theta)-1)}{sin(\theta)(2sin(\theta)-1)}

= t a n − 1 c o s ( θ ) s i n ( θ ) =tan^{-1}\frac{cos(\theta)}{sin(\theta)}

= t a n − 1 ( c o t ( θ ) =tan^{-1}(cot(\theta)

= t a n − 1 ( t a n ( π 2 − θ ) =tan^{-1}(tan(\frac{\pi}{2}-\theta)

= 90 − θ =\boxed{90-\theta}

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