Question

What is the maximum value of the function \[a\cos x + b\sin x\]?
A. \[a + b\]
B. \[a - b\]
C. \[\left| a \right| + \left| b \right|\]
D. \[{\left( {{a^2} + {b^2}} \right)^{\dfrac{1}{2}}}\]

Answer 1

Hint Simplify the given function by multiplying the numerator and denominator by \[\sqrt {\left( {{a^2} + {b^2}} \right)} \]. Consider \[\dfrac{a}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }} = \cos \theta \] and \[\dfrac{b}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }} = \sin \theta \] then further simplify the function in the form \[\theta \]. In the end, substitute the maximum value of \[\cos \] and get the maximum value of the function.

Formula used
\[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]
The range of the trigonometric function \[\cos\theta \] is \[\left[ { - 1,1} \right]\].

Complete step by step solution:
The given function is \[a\cos x + b\sin x\].
Let consider,
\[f\left( x \right) = a\cos x + b\sin x\]
Now multiply the numerator and denominator of the right-hand side of the above function by \[\sqrt {\left( {{a^2} + {b^2}} \right)} \].
\[f\left( x \right) = \sqrt {\left( {{a^2} + {b^2}} \right)} \left[ {\dfrac{a}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}\cos x + \dfrac{b}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}\sin x} \right]\]
Now consider \[\cos \theta = \dfrac{a}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}\] and \[\sin \theta = \dfrac{b}{{\sqrt {\left( {{a^2} + {b^2}} \right)} }}\].
\[f\left( x \right) = \sqrt {\left( {{a^2} + {b^2}} \right)} \left[ {\cos \theta \cos x + \sin \theta \sin x} \right]\]
Now apply the reverse formula \[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\].
\[f\left( x \right) = \sqrt {\left( {{a^2} + {b^2}} \right)} \left[ {\cos\left( {\theta - x} \right)} \right]\]
The maximum value of trigonometric function \[\cos\] is 1.
So, the maximum value of \[\cos\left( {\theta - x} \right)\] is 1.
Thus, the maximum value of \[f\left( x \right)\] is \[f\left( x \right) = \sqrt {\left( {{a^2} + {b^2}} \right)} \].
\[f\left( x \right) = \sqrt {\left( {{a^2} + {b^2}} \right)} = {\left( {{a^2} + {b^2}} \right)^{\dfrac{1}{2}}}\]
Hence the correct option is D.

Note: Student often confused with the formula \[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\] and \[\cos \left( {A - B} \right) = \cos A\cos B - \sin A\sin B\]. The correct formula is \[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\].