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Solution to puzzle 109: Nested circular functions

Let x be a real number.  Which is greater, sin(cos x) or cos(sin x)?


We will make use of the following trigonometric identities:

sin A = cos(pi/2 − A)(1)
cos A − cos B = −2 sin(½(A + B)) sin(½(A − B))(2)
A cos x + B sin x = R cos (x − c),  where R = square root(A2 + B2), and c = arctan(B/A)  (3)

Applying the above, we obtain

cos(sin x) − sin(cos x) = cos(sin x) − cos(pi/2 − cos x),  from (1).
  = −2 sin[½(sin x − cos x + pi/2)] sin[½(sin x + cos x − pi/2)],  from (2).
  = −2 sin(−½root 2cos(x + pi/4) + pi/4) sin(½root 2cos(x − pi/4) − pi/4),  from (3).

Since pi/4 > ½root 2,  0 < −½root 2cos(x + pi/4) + pi/4 < pi/2, for all x.
Hence sin(−½root 2cos(x + pi/4) + pi/4) > 0, for all x.
Similarly, −pi/2 < ½root 2cos(x − pi/4) − pi/4 < 0, for all x.
Hence sin(½root 2cos(x − pi/4) − pi/4) < 0, for all x.
It follows that −2 sin(−½root 2cos(x + pi/4) + pi/4) sin(½root 2cos(x − pi/4) − pi/4) > 0.

Therefore cos(sin x) > sin(cos x) for all real x.


Remarks

A proof using periodicity is given on this wu :: forums thread by Eigenray.  (Select the hidden text in order to view it.)

A sketch of the graph of y = cos(sin x) − sin(cos x) is given below.  The graph has period 2pi.

Graph of y = cos(sin x) - sin(cos x)

Source: Examples of Problems (page since taken down)

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