Angle of displacement

Geometry Level 3

A stair case has the property such that it starts at point A ( 0 , 0 ) A (0,0) and travels in the pattern right, up, right, up, right, up. . . ad infinitum and ends up at point B ( x , y ) B(x,y) .

It travels cos ⁡ ( 1 8 ∘ ) \cos\left(18^\circ \right) to the right, cos ⁡ 2 ( 1 8 ∘ ) \cos^2\left(18^\circ \right) up, cos ⁡ 3 ( 1 8 ∘ ) \cos^3\left(18^\circ \right) right, cos ⁡ 4 ( 1 8 ∘ ) \cos^4\left(18^\circ \right) and so on indefinitely.

Find the angle θ \theta segment A B ‾ \overline{AB} forms with the x x -axis in degrees.

Assume θ < 9 0 ∘ \theta<90^{\circ} .


The answer is 43.56.

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Trevor Arashiro by Trevor Arashiro , 22, USA

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4 solutions

Rohit Sachdeva
May 12, 2015

Why cant we just see that at each step we will have y=(cos18°)x

(cos18°,cos²18°) ;

((cos18°+cos³18°),(cos²18°+cos⁴18°))

and so on....

So points A & B will lie on the line y=(cos18°)x

Hence slope tan∅=cos18°

Or ∅=arctan(cos18°) =43.56°

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Moderator note:

Yes, this is the simplest of all. You just need to show that the points lie on a straight line and voilà you got the answer! Very intuitive. Wonderful.

I like this solution the most! Well done.

Jake Lai - 6 years, 1 month ago

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I agree most neat way I can say! Well done

shivamani patil - 6 years, 1 month ago
Chew-Seong Cheong
May 11, 2015

A ( 0 , 0 A(0,0 and B ( x , y ) B(x,y) , therefore,

x = ∑ k = 1 ∞ cos ⁡ 2 k − 1 ( π 10 ) = ∑ k = 0 ∞ cos ⁡ k ( π 10 ) − ∑ k = 0 ∞ cos ⁡ 2 k ( π 10 ) = 1 1 − cos ⁡ ( π 10 ) − 1 1 − cos ⁡ 2 ( π 10 ) = cos ⁡ ( π 10 ) sin ⁡ 2 ( π 10 ) y = ∑ k = 1 ∞ cos ⁡ 2 k ( π 10 ) = ∑ k = 0 ∞ cos ⁡ 2 k ( π 10 ) − 1 = 1 1 − cos ⁡ 2 ( π 10 ) − 1 = cos ⁡ 2 ( π 10 ) sin ⁡ 2 ( π 10 ) tan ⁡ θ = y x = cos ⁡ 2 ( π 10 ) sin ⁡ 2 ( π 10 ) × sin ⁡ 2 ( π 10 ) cos ⁡ ( π 10 ) = cos ⁡ ( π 10 ) \begin{aligned} x & = \sum_{k=1}^\infty {\cos^{2k-1}{\left( \frac{\pi}{10} \right)}} = \sum_{k=0}^\infty {\cos^k {\left( \frac{\pi}{10} \right)}} - \sum_{k=0}^\infty {\cos^{2k} {\left( \frac{\pi}{10} \right)}} \\ & = \frac{1}{1-\cos {\left( \frac{\pi}{10} \right)}} - \frac{1}{1-\cos^{2} {\left( \frac{\pi}{10} \right)}} = \frac{\cos {\left( \frac{\pi}{10} \right)}}{\sin^2 {\left( \frac{\pi}{10} \right)}} \\ y & = \sum_{k=1}^\infty {\cos^{2k}{\left( \frac{\pi}{10} \right)}} = \sum_{k=0}^\infty {\cos^{2k}{\left( \frac{\pi}{10} \right)}} - 1 \\ & = \frac{1}{1-\cos^{2} {\left( \frac{\pi}{10} \right)}} - 1 = \frac{\cos^{2} {\left( \frac{\pi}{10} \right)}}{\sin^{2} {\left( \frac{\pi}{10} \right)}} \\ \tan{\theta} & = \frac {y}{x} = \frac{\cos^{2} {\left( \frac{\pi}{10} \right)}}{\sin^{2} {\left( \frac{\pi}{10} \right)}} \times \frac{\sin^{2} {\left( \frac{\pi}{10} \right)}}{\cos {\left( \frac{\pi}{10} \right)}} = \cos {\left( \frac{\pi}{10} \right)} \end{aligned}

⇒ θ = tan ⁡ − 1 cos ⁡ ( π 10 ) = 43.5 6 ∘ \Rightarrow \theta = \tan^{-1} {\cos {\left( \frac{\pi}{10} \right)}} = \boxed{43.56^\circ} to 2 decimal places.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

You should be consistent with the measurement of angles you used. Since you started with radians, it's better to end your solution with " … = tan ⁡ − 1 cos ⁡ ( π 10 ) = 0.7603 \ldots = \tan^{-1} {\cos {\left( \frac{\pi}{10} \right)}} = 0.7603 which is equivalent to 43.5 6 ∘ 43.56^\circ degrees." Nonetheless, a very conventional approach, and a very neat solution as usual. Nicely done.

The net distance traveled to the right is

X = ∑ k = 1 ∞ ( − 1 ) k + 1 ( cos ⁡ ( π 10 ) ) 2 k − 1 = cos ⁡ ( π 10 ) 1 + cos ⁡ 2 ( π 10 ) . X = \displaystyle\sum_{k=1}^{\infty} (-1)^{k+1}(\cos(\frac{\pi}{10}))^{2k-1} = \dfrac{\cos(\frac{\pi}{10})}{1 + \cos^{2}(\frac{\pi}{10})}.

The net distance traveled up is

Y = ∑ k = 1 ∞ ( − 1 ) k + 1 ( cos ⁡ ( π 10 ) ) 2 k = cos ⁡ 2 ( π 10 ) 1 + cos ⁡ 2 ( π 10 ) . Y = \displaystyle\sum_{k=1}^{\infty} (-1)^{k+1} (\cos(\frac{\pi}{10}))^{2k} = \dfrac{\cos^{2}(\frac{\pi}{10})}{1 + \cos^{2}(\frac{\pi}{10})}.

The desired angle is then

θ = arctan ⁡ ( Y X ) = arctan ⁡ ( cos ⁡ ( π 10 ) ) = 43.5 6 ∘ \theta = \arctan\left(\dfrac{Y}{X}\right) = \arctan(\cos(\frac{\pi}{10})) = \boxed{43.56^{\circ}} to 2 decimal places.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

You should be consistent with the measurement of angles you used. Since you started with radians, it's better to end your solution with " … = tan ⁡ − 1 cos ⁡ ( π 10 ) = 0.7603 \ldots = \tan^{-1} {\cos {\left( \frac{\pi}{10} \right)}} = 0.7603 which is equivalent to 43.5 6 ∘ 43.56^\circ degrees." Nonetheless, your solution is quicker than of Chew-Seong Cheong's solution. Neat trick!

Janine Yu
May 25, 2015

with the conditions given, the length we have to the right= cos18 + cos^3(18)+cos^5(18)+cos^7(18)+.... the length we have upward= cos^2(18)+cos^4(18)+cos^6(18)+cos^8(18)+.....

to solve for the sum of an infinite geometric sequence, we use the formula S= (first term/(1-r))

Hence, length to the right= (cos18/(1-cos^2(18))) length upwards= (cos^2(18)/(1-cos^2(18)))

With that, we'll form a right triangle with (cos18/(1-cos^2(18))) as the length of the adjacent side and (cos^2(18)/(1-cos^2(18))) as the length of the opposite side in accordance to angle (theta).

Therefore, using trigonometric ratio of tangent, tan(theta)= cos18 theta= tan^-1(cos18) = 43.56

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