How's this related to trig?

Calculus Level 5

$\large \dfrac{7}{5} = x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} - \dfrac{x^4}{4!} + \dfrac{x^5 }{5!} + \dfrac{x^6}{6!} - \dfrac{x^7}{7!} - \dfrac{x^8}{8!} + ...$

If the value of $x$ that satisfies the above equation can be expressed in the form $2 \arctan \left( \dfrac{-a + \sqrt{b}}{c}\right) + 2\pi k$ for integer $k$ , where $a$ and $c$ are positive coprime integers and $b$ is square-free, find $a + b + c$ .

The answer is 54.

1 solution

Michael Mendrin
May 7, 2016

This expression can be restated in terms of series expansions of familiar trigonometric functions

$\dfrac { 7 }{ 5 } =1+Sin\left( x \right) -Cos\left( x \right)$

If we let

$x=2ArcTan\left(y\right)+2\pi k$

then we have though substitutions

$\dfrac { 7 }{ 5 } =1+\dfrac { 2y }{ { y }^{ 2 }+1 } -\dfrac { -{ y }^{ 2 }+1 }{ { y }^{ 2 }+1 }$

Solving for $y$ gets us

$y=\dfrac { -5\pm \sqrt { 46 } }{ 3 }$

and we have our answer $54$

Hi can you explain how you substituted x=2tan^-1(y) +2pik into sin(x) and cos(x) to get 2y/(y^2 +1) and (-y^2+1)/(y^2+1)

Ashish Sacheti - 5 years, 1 month ago

How's this related to trig?

The answer is 54.

1 solution

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