# 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(/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 = (A2 + B2), and c = arctan(B/A) (3)

Applying the above, we obtain

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

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

Source: Examples of Problems (page since taken down)