Tangents and Cotangents

Calculus Level 5

Graph showing region inside r = 1 + tan ( 2 θ ) r = 1 + \tan(2\theta) and r = 1 + cot ( 2 θ ) r = 1 + \cot(2\theta) .

Graph showing region inside r = 1 + tan ( 3 θ ) r = 1 + \tan(3\theta) and r = 1 + cot ( 3 θ ) r = 1 + \cot(3\theta) .

Let n n be a positive integer 2 \geq 2 .

Find the area inside r = 1 + tan ( n θ ) r = 1 + \tan(n\theta) and r = 1 + cot ( n θ ) r = 1 + \cot(n\theta) .


The answer is 4.

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1 solution

Rocco Dalto
Jul 6, 2018

Let n n be a positive integer and n 2 n \geq 2 .

The area A = n ( 3 π 4 n 5 π 4 n ( 1 + tan ( n θ ) ) 2 d θ + 5 π 4 n 7 π 4 n ( 1 + cot ( n θ ) ) 2 d θ ) = A = n(\displaystyle\int_{\frac{3\pi}{4n}}^{\frac{5\pi}{4n}} (1 + \tan(n\theta))^2 d\theta + \displaystyle\int_{\frac{5\pi}{4n}}^{\frac{7\pi}{4n}} (1 + \cot(n\theta))^2 d\theta) =

n ( 3 π 4 n 5 π 4 n 2 tan ( n θ ) + sec 2 ( n θ ) d θ + 5 π 4 n 7 π 4 n 2 cot ( n θ ) + csc 2 ( n θ ) d θ ) = n(\displaystyle\int_{\frac{3\pi}{4n}}^{\frac{5\pi}{4n}} 2\tan(n\theta) + \sec^2(n\theta) d\theta + \displaystyle\int_{\frac{5\pi}{4n}}^{\frac{7\pi}{4n}} 2\cot(n\theta) + \csc^2(n\theta) d\theta) =

n ( ( 2 n ln cos ( n θ ) + 1 n tan ( n θ ) ) 3 π 4 n 5 π 4 n + ( 2 n ln sin ( n θ ) 1 n cot ( n θ ) ) 5 π 4 n 7 π 4 n ) = n((-\dfrac{2}{n}\ln|\cos(n\theta)| + \dfrac{1}{n}\tan(n\theta))|_{\frac{3\pi}{4n}}^{\frac{5\pi}{4n}} + (\dfrac{2}{n}\ln|\sin(n\theta) - \dfrac{1}{n}\cot(n\theta))|_{\frac{5\pi}{4n}}^{\frac{7\pi}{4n}}) = n ( 4 n ) = 4 n(\dfrac{4}{n}) = \boxed{4} .

You can use \displaystyle for integrals like this 3 π 4 n 5 π 4 n \displaystyle \int_{\frac{3\pi}{4n}}^\frac{5\pi}{4n}

X X - 2 years, 11 months ago

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Thanks. Much better.

Rocco Dalto - 2 years, 11 months ago

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