Trigonometry
Geometry
Level
3
What is the maximum possible value of 2997.sin x+3996.cosx
The answer is 4995.
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1 solution
The maximum value of (Asinx + Bcosx) is root(A^2+B^2).
Define an angle φ in the following way:
sinφ = b/√(a² + b²) and cosφ = a/√(a² + b²).
Note that it is perfectly legitimate to do this because
sin²φ + cos²φ = (b² + a²)/(a² + b²) = 1.
The above has to be true of any angle φ that can exist!
f(x) = √(a² + b²) [a/√(a² + b²) sin(x) + b/√(a² + b²) cos(x)].
But the coefficients are the sine and cosine we just defined. So
f(x) = √(a² + b²) [cosφ sin(x) + sinφ cos(x)]
Using the sum of angles formula for the sine function, this simplifies to
f(x) = √(a² + b²) sin(x + φ).
In this form, its easy to see that the amplitude of f is √(a² + b²). So the maximum and minimum values are just this and its negative.
Bonus - Can anyone prove it using the much graceful Cauchy Schwarz inequality?