System of tan equations

Geometry Level 4

{ x + y + z = 3 π / 4 tan ⁡ ( x ) + tan ⁡ ( y ) + tan ⁡ ( z ) = 5 tan ⁡ ( x ) tan ⁡ ( y ) tan ⁡ ( z ) = 1 \large{\begin{cases} x+y+z=3\pi/4 \\ \tan( x)+\tan (y)+\tan (z)=5 \\ \tan (x)\tan (y) \tan( z)=1 \\ \end{cases} }

x , y , z x,y,z are real numbers satisfying the system of equations above Find the value of tan ⁡ 3 ( x ) + tan ⁡ 3 ( y ) + tan ⁡ 3 ( z ) \tan^3( x) + \tan^3( y) +\tan^3 (z ) .


The answer is 53.

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Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

Ravi Dwivedi
Jul 9, 2015

Using the formula tan ⁡ ( x + y + z ) = tan ⁡ x + tan ⁡ y + tan ⁡ z − tan ⁡ x ⋅ tan ⁡ y ⋅ tan ⁡ z 1 − tan ⁡ x ⋅ tan ⁡ y − tan ⁡ y ⋅ tan ⁡ z − tan ⁡ z ⋅ tan ⁡ x \tan(x+y+z)=\frac{\tan x+\tan y+\tan z -\tan x \cdot \tan y \cdot \tan z}{1-\tan x \cdot \tan y - \tan y \cdot \tan z - \tan z \cdot \tan x} \\ we have

− 1 = tan ⁡ 3 π 4 = 5 − 1 1 − tan ⁡ x ⋅ tan ⁡ y − tan ⁡ y ⋅ tan ⁡ z − tan ⁡ z ⋅ tan ⁡ x ⟹ tan ⁡ x ⋅ tan ⁡ y + tan ⁡ y ⋅ tan ⁡ z + tan ⁡ z ⋅ tan ⁡ x = 5 -1=\tan \frac{3\pi}{4}=\frac{5-1}{1-\tan x \cdot \tan y - \tan y \cdot \tan z - \tan z \cdot \tan x}\\ \implies \tan x \cdot \tan y + \tan y \cdot \tan z + \tan z \cdot \tan x =5 \\ Using Vieta's relations we conclude that tan ⁡ x , tan ⁡ y , tan ⁡ z \tan x,\tan y, \tan z are roots of the polynomial t 3 − 5 t 2 + 5 t − 1 = 0 tan ⁡ 3 ( x ) = 5 tan ⁡ 2 ( x ) − 5 tan ⁡ x + 1 tan ⁡ 3 ( y ) = 5 tan ⁡ 2 ( y ) − 5 tan ⁡ y + 1 tan ⁡ 3 ( z ) = 5 tan ⁡ 2 ( z ) − 5 tan ⁡ z + 1 t^3-5t^2+5t-1=0 \\ \tan^3( x)=5\tan^2( x)-5\tan x+1 \\ \tan^3( y)=5\tan^2( y)-5\tan y+1 \\ \tan^3( z)=5\tan^2( z)-5\tan z+1 \\

Adding these three equations we get

tan ⁡ 3 ( x ) + tan ⁡ 3 ( y ) + tan ⁡ 3 ( z ) = 5 ( tan ⁡ 2 ( x ) + tan ⁡ 2 ( y ) + tan ⁡ 2 ( z ) ) − 5 ( tan ⁡ x + tan ⁡ y + tan ⁡ z ) + 3 \tan^3( x)+\tan^3(y)+\tan^3(z) = 5(\tan^2( x)+\tan^2( y)+\tan^2( z))-5(\tan x+\tan y+\tan z)+3

tan ⁡ 3 ( x ) + tan ⁡ 3 ( y ) + tan ⁡ 3 ( z ) = 5 ( ( tan ⁡ x + tan ⁡ y + tan ⁡ z ) 2 − 2 ( tan ⁡ x ⋅ tan ⁡ y + tan ⁡ y ⋅ tan ⁡ z + tan ⁡ z ⋅ tan ⁡ x ) ) − 5 ( tan ⁡ x + tan ⁡ y + tan ⁡ z ) + 3 \tan^3( x)+\tan^3(y)+\tan^3(z) = 5((\tan x+\tan y+\tan z)^2 -2(\tan x \cdot \tan y + \tan y \cdot \tan z + \tan z \cdot \tan x))-5(\tan x+\tan y+\tan z)+3

tan ⁡ 3 ( x ) + tan ⁡ 3 ( y ) + tan ⁡ 3 ( z ) = 5 ( 5 2 − 2 ( 5 ) ) − 5 ( 5 ) + 3 = 53 \tan^3( x)+\tan^3(y)+\tan^3(z) =5(5^2-2(5))-5(5)+3=\boxed{53}

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Moderator note:

There isn't a need to use Vieta's.

We know that a 3 + b 3 + c 3 = ( a + b + c ) 3 − 3 ( a + b + c ) ( a b + b c + c a ) + 3 a b c a^3 + b^3 + c^3 = ( a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc (which is a "well-known algebraic identity", which also follows from Newton's identities).

Yeah U r right.....but why I did that unconsciously because I thought that way when I posed the problem

Ravi Dwivedi - 5 years, 11 months ago

I am getting the answer as 68.please point out the mistake. tan ⁡ ( x + y ) = − 1 − tan ⁡ z 1 − t a n z ⟹ ( tan ⁡ x + tan ⁡ y ) = 1 + tan ⁡ z \tan(x+y)=\dfrac{-1-\tan z}{1-tan z}\\ \Longrightarrow (\tan x+\tan y)\ =1+\tan z ,similarly we would get three equations like this.Adding the three we would get the value of \ ∑ c y c tan ⁡ x tan ⁡ y = 4 \ \sum_{cyc}^{}\tan x \tan y=4 . From this we would get tan ⁡ 2 x + tan ⁡ 2 y + tan ⁡ 2 z = 17 \tan ^2 x+\tan^2 y+\tan ^2 z=17 After substituting in the identity ( a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) (a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) we get the value of our required expression 68. @Ravi Dwivedi @Calvin Lin

satyendra kumar - 5 years, 10 months ago

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How did you get from the first line to the second?

I also do not see how "adding the three gives us tan ⁡ x tan ⁡ y = 4 \tan x \tan y = 4 ". If anything, I get ∑ tan ⁡ x = 1 \sum \tan x = 1 (which contradicts a given condition).

Calvin Lin Staff - 5 years, 10 months ago

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