Academy — Sanjib Sikder

sin(cos x) is not a mess. Establish the range of the inner function first — everything else follows. The technique JEE Advanced uses repeatedly.

What does sin(cos x) look like? Before drawing it, answer a prior question: what values does cos x take? Only values in [−1, 1]. So sin(cos x) is sin applied to inputs restricted to [−1, 1].

Range analysis before graph shape

For y = sin(cos x):

cos x ∈ [−1, 1] always
So the argument of the outer sin is always in [−1, 1]
sin is increasing on [−π/2, π/2], and [−1, 1] ⊂ [−π/2, π/2]
So y = sin(cos x) ∈ [sin(−1), sin(1)] = [−sin 1, sin 1] ≈ [−0.841, 0.841]

The curve never reaches ±1. Its effective amplitude is sin(1) ≈ 0.841.

Period of sin(cos x)

cos x has period 2π. So sin(cos x) also has period 2π — the outer function doesn't change the period of the inner one.

y = cos(sin x)

sin x ∈ [−1, 1]
cos is even, so cos(sin x) = cos(|sin x|) — the curve is always even
cos on [−1, 1] gives values in [cos 1, 1] ≈ [0.540, 1]
So cos(sin x) is always positive, oscillates between cos(1) and 1

The curve stays in the upper half of the plane entirely. Period 2π. The "amplitude" is (1 − cos 1)/2 ≈ 0.23.

The JEE question this sets up

Does sin(cos x) = cos(sin x) have solutions?

sin(cos x) can be negative (since sin(cos x) ∈ [−sin 1, sin 1]). cos(sin x) is always positive. So the two curves can only intersect where sin(cos x) ≥ 0, i.e., where cos x ≥ 0.

On [0, π]: cos x ≥ 0 only on [0, π/2]. Check if sin(cos x) = cos(sin x) at any point there — numerically, there are no exact solutions, and this can be shown by monotonicity arguments on [0, π/2].

The key move in every composite problem: establish the range of the inner function first. The outer function operates on that restricted set. Everything else follows.

General strategy for composites

Given f(g(x)):

Find the range of g(x) — call it I
Restrict f to domain I
Find the range of f on I — that is the range of the composite
The period of f(g(x)) equals the period of g(x)
Draw the graph by substituting key values of g(x)

Common composites in JEE

Function	Range of inner	Range of composite
sin(cos x)	[−1, 1]	[−sin 1, sin 1] ≈ [−0.84, 0.84]
cos(sin x)	[−1, 1]	[cos 1, 1] ≈ [0.54, 1]
sin(sin x)	[−1, 1]	[−sin 1, sin 1] ≈ [−0.84, 0.84]
tan(sin x)	[−1, 1]	[−tan 1, tan 1] ≈ [−1.56, 1.56]
sin(tan x)	ℝ	[−1, 1] — same as ordinary sin