Let x1 and x2 be the roots of the equation : x2−22x+1=0Calculate arctan⁡(x1)⋅arctan⁡(x2)

We havex2−22x+1=(x−(1+2))(x−(−1+2))=0so let x1=1+2 and x2=−1+2Let a=tan−1(1+2). Using the double angle tangent formula,tan⁡2a=2tan⁡a1−tan2a⇒tan⁡2a=2(1+2)1−(1+2)2=−1and hence taking the principal angle givestan−1(1+2)=38πLet b=tan−1(−1+2). Using the double angle tangent formula,tan⁡2b=2tan⁡b1−tan2b⇒tan⁡2b=2(−1+2)1−(−1+2)2=1and hence taking the principal angle givestan−1(−1+2)=18πThereforetan−1(x1)⋅tan−1(x2)=38π⋅18π=364π2as desired.

Expert answers to "Let x1 and x2 be the..."

London Douglas

Answered question

2022-03-21

Let

x_{1} and x_{2}

be the roots of the equation :

x^{2} - 2 \sqrt{2} x + 1 = 0

Calculate

\arctan (x_{1}) \cdot \arctan (x_{2})

Answer & Explanation

Cason Singleton

Beginner2022-03-22Added 13 answers

We have

x^{2} - 2 \sqrt{2} x + 1 = (x - (1 + \sqrt{2})) (x - (- 1 + \sqrt{2})) = 0

so let

x_{1} = 1 + \sqrt{2}

and

x_{2} = - 1 + \sqrt{2}

Let

a = \tan^{- 1} (1 + \sqrt{2})

. Using the double angle tangent formula,

\tan 2 a = \frac{2 \tan a}{1 - \tan^{2} a} \Rightarrow \tan 2 a = \frac{2 (1 + \sqrt{2})}{1 - {(1 + \sqrt{2})}^{2}} = - 1

and hence taking the principal angle gives

\tan^{- 1} (1 + \sqrt{2}) = \frac{3}{8} π

Let

b = \tan^{- 1} (- 1 + \sqrt{2})

. Using the double angle tangent formula,

\tan 2 b = \frac{2 \tan b}{1 - \tan^{2} b} \Rightarrow \tan 2 b = \frac{2 (- 1 + \sqrt{2})}{1 - {(- 1 + \sqrt{2})}^{2}} = 1

and hence taking the principal angle gives

\tan^{- 1} (- 1 + \sqrt{2}) = \frac{1}{8} π

Therefore

\tan^{- 1} (x_{1}) \cdot \tan^{- 1} (x_{2}) = \frac{3}{8} π \cdot \frac{1}{8} π = \frac{3}{64} π^{2}

as desired.

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