Squared Trig

Do not draw sin²x from scratch. Convert it first.

$\sin^2 x = \frac{1 - \cos 2x}{2}$

Read this as: amplitude ½, frequency doubled (period π), midline at y = ½, and a reflection (the minus sign before cos 2x).

In the four-parameter form: A = −½, B = 2, C = 0, D = ½.

The explorer below is pre-seeded to match sin²x. Confirm the period is π and the midline is y = 0.5.

What the graph looks like

• Oscillates between 0 and 1. Never negative (it's a square).
• Period is π, not 2π. This catches students in MCQs.
• The peaks are at x = π/2 + nπ (where sin x = ±1, so sin²x = 1).
• The zeros are at x = nπ (where sin x = 0).

cos²x

$\cos^2 x = \frac{1 + \cos 2x}{2}$

Same amplitude, same period π, same midline y = ½. But it peaks at x = 0, nπ and zeros at x = π/2 + nπ. It is sin²x shifted by π/2.

Note: sin²x + cos²x = 1 is visible in the graphs — at every x, the two curves add to 1, which means they are reflections of each other about the midline y = ½.

sin x · cos x

$\sin x \cos x = \frac{\sin 2x}{2}$

Period π, amplitude ½, no vertical shift. This is a standard sin curve — just compressed and scaled. Not a new shape at all.

Why this matters for JEE

JEE integration and differentiation problems routinely involve sin²x and cos²x. Students who know the double-angle form write down the answer in one step. Students who try to handle sin²x directly waste time.

Also: the period of sin²x is π, not 2π. Asking "what is the period of sin²x + cos²x?" is a trap — sin²x + cos²x = 1 is a constant, period undefined (or arbitrary).

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Next: Solution counting — using graphs to answer "how many solutions" questions in seconds.