cos of sin OR sin of cos

Level 2

$\color{#3D99F6}{A = \cos{(\sin{(\cos{(\sin{(\cos{(\sin{(\ldots)})})})})})} }$ $\color{#D61F06}{B = \sin{(\cos{(\sin{(\cos{(\sin{(\cos{(\ldots)})})})})})}}$

If this alternates between $\sin$ and $\cos$ infinitely many times, which one will be greater?

Inspiration: here

\color{#3D99F6}{A}

\color{#D61F06}{B}

A=B

1 solution

Sachetan Debray
Jun 1, 2020

All right, I am posting a solution without graphics or pictures. Please try to visualize as you go along. If you want me to use pictures, please say so in the comments box and I will edit my answer.(I don't yet know where to drag in animation or graphics from, so please mention that as well)

By substitution, A = cos B

B = sin A

Visualize the graph of sine curve. Take a point (A,0) on the x-axis.

B=sin A

So imagine a point P on the graph at distance $\sqrt{ A^2 + B^2}$ from the origin, with ordinate=B and abscissa=A.

Visualize the graph of cosine curve. Take a point (B,0) on the x-axis.

A=cos B

So imagine a point Q on the graph at distance $\sqrt{ A^2 + B^2}$ from the origin, with ordinate=A and abscissa=B

Superimpose the graph of cosine on the graph of sine curve. It is clear that points P and Q will coincide since they are at the same distance from the origin.

Thus the ordinate of point P=ordinate of point Q

Thus B=A

Note-I am assuming that A and B are both less than 90 degrees and more than 0 degrees, as that was the condition in the link given for the inspiration for this problem.

But if my assumption is false, do let me know in the comments below.

I think it can be proved for the other cases as well (where A, B are less than 0 degrees or more than 90 degrees), though it may take some time.

You can use WolphramAlpha(PC) or GeoGebra(Mobile) :)

Jeff Giff - 1 year ago

Do I have to download it separately?(I have Microsoft OS in my PC)

Sachetan Debray - 1 year ago

cos of sin OR sin of cos

1 solution

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