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MathsMediumClass 11

Maximum & Minimum Values — a sinθ + b cosθ

Trigonometric Functions & Equations

8

JEE Qs

8%

Hard

35

min

Always transform 'a sinθ + b cosθ' into 'R sin(θ ± α)' or 'R cos(θ ± α)' using R = sqrt(a^2 + b^2) to quickly find its range.

🧮 Key Formulas

Maximum value of a sinθ + b cosθ is sqrt(a^2 + b^2)
Minimum value of a sinθ + b cosθ is -sqrt(a^2 + b^2)
Range of a sinθ + b cosθ is [-sqrt(a^2 + b^2), sqrt(a^2 + b^2)]
a sinθ + b cosθ = R sin(θ + α), where R = sqrt(a^2 + b^2), cosα = a/R, sinα = b/R
a sinθ + b cosθ = R cos(θ - β), where R = sqrt(a^2 + b^2), cosβ = b/R, sinβ = a/R

✅ Key Points for JEE

  • 1The core idea is to transform the expression a sinθ + b cosθ into a single trigonometric ratio, R sin(θ ± α) or R cos(θ ± α).
  • 2The maximum and minimum values are directly obtained from the range of the sine/cosine function, which is [-1, 1].
  • 3This method provides an elegant way to find extreme values without using calculus, relying solely on trigonometric identities.
  • 4The phase angle α or β determines the angle θ at which the maximum or minimum occurs, but not the max/min value itself.

⚠️ Common Mistakes

  • Incorrectly assuming the maximum value is |a| + |b| or a+b, and minimum value is a-b.
  • Errors in calculating R = sqrt(a^2 + b^2), such as forgetting to square 'a' and 'b' or omitting the square root.
  • Failing to recognize the a sinθ + b cosθ form when it appears embedded within a more complex function or expression.

NCERT Chapters

  • Class 11 Mathematics Ch 3: Trigonometric Functions