All subjects · Mathematics · WMA13 (P3)

Differentiating sin, cos and tan

The derivatives of the trig functions are short to learn. They only work when the angle is measured in radians.

The results

Sums and differences are differentiated term by term, along with ekx and ln x (see the lesson on differentiating ex and ln x).

Why radians?

The results come from the fact that sin h/h gets close to 1 as h gets small, which is only true in radians. In degrees, extra factors of π/180 appear. Set your calculator to radians for these questions.

Using them

Tangents, normals and stationary points work as in P1 and P2: find dy/dx, substitute the x-value, then use y - y1 = m(x - x1). For a normal, use gradient -1/m.

Worked example

Differentiate y = 3sin 2x - cos 4x.

  1. d/dx (3sin 2x) = 3 × 2cos 2x = 6cos 2x
  2. d/dx (-cos 4x) = -(-4sin 4x) = 4sin 4x

Answer: dy/dx = 6cos 2x + 4sin 4x

Key idea: sin kx gives k cos kx; cos kx gives -k sin kx; tan kx gives k sec2 kx. Always work in radians.

Check yourself

  1. Differentiate cos 5x.
    • 5sin 5x
    • -5sin 5x
    • -sin 5x
    • -(1/5)sin 5x
    Show answer

    -5sin 5x. d/dx (cos kx) = -k sin kx with k = 5.

  2. Find the gradient of y = sin 3x at x = π/9.
    Show answer

    1.5. dy/dx = 3cos 3x. At x = π/9: 3cos(π/3) = 3 × 1/2 = 1.5.

  3. Find the gradient of y = tan 2x at x = π/8.
    Show answer

    4. dy/dx = 2sec2 2x. At x = π/8: 2sec2(π/4) = 2 × 2 = 4.

  4. Find the gradient of y = e2x + sin 4x at x = 0.
    Show answer

    6. dy/dx = 2e2x + 4cos 4x. At x = 0: 2 + 4 = 6.

  5. Why must x be in radians when you use d/dx (sin x) = cos x?
    • The result relies on sin h/h getting close to 1, which is only true in radians
    • Calculators cannot find sin in degrees
    • Degrees are not allowed in A level
    • sin x is undefined in degrees
    Show answer

    The result relies on sin h/h getting close to 1, which is only true in radians. In degrees, sin h/h tends to π/180, so an extra factor appears.

  6. Find the x-coordinate of the stationary point of y = x + 2cos x for 0 < x < π/2, to 3 decimal places.
    Show answer

    0.524. dy/dx = 1 - 2sin x = 0, so sin x = 1/2 and x = π/6 = 0.524.

  7. Find the gradient of the normal to y = 3cos 2x at x = π/4. Give your answer as a fraction in its simplest form.
    Show answer

    1/6. dy/dx = -6sin 2x. At x = π/4: -6sin(π/2) = -6. The normal gradient is -1/(-6) = 1/6.

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Last updated 7 November 2026. Spotted a mistake? Tell us.