Question

Find the point where \[\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {x + \dfrac{\pi }{6}} \right)\] is attained maxima in the interval \[\left( {0,\dfrac{\pi }{2}} \right)\].
A. \[x = \dfrac{\pi }{{12}}\]
B. \[x = \dfrac{\pi }{6}\]
C. \[x = \dfrac{\pi }{3}\]
D. \[x = \dfrac{\pi }{2}\]

Answer 1

Hint:First we will multiply and divide \[\sqrt 2 \] with given expression \[\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {x + \dfrac{\pi }{6}} \right)\]. Then replace \[\dfrac{1}{{\sqrt 2 }}\] by \[\cos \dfrac{\pi }{4}\] in the first term and \[\dfrac{1}{{\sqrt 2 }}\] by \[\sin \dfrac{\pi }{4}\] in the second term. Then we will apply the formula \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\]. We know that the maxima value of \[\sin \theta \] is 1. By using this concept, we will calculate the point where \[\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {x + \dfrac{\pi }{6}} \right)\] is maxima.

Formula Used:
\[\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
\[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\]
The maxima value of \[\sin \theta \] is 1.

Complete step by step solution:
Given trigonometry function is \[\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {x + \dfrac{\pi }{6}} \right)\]
Now multiply and divide \[\sqrt 2 \] with given expression
\[ = \sqrt 2 \left[ {\dfrac{1}{{\sqrt 2 }}\sin \left( {x + \dfrac{\pi }{6}} \right) + \dfrac{1}{{\sqrt 2 }}\cos \left( {x + \dfrac{\pi }{6}} \right)} \right]\]
Replace \[\dfrac{1}{{\sqrt 2 }}\] by \[\cos \dfrac{\pi }{4}\] in the first term and \[\dfrac{1}{{\sqrt 2 }}\] by \[\sin \dfrac{\pi }{4}\] in the second term
\[ = \sqrt 2 \left[ {\cos \dfrac{\pi }{4}\sin \left( {x + \dfrac{\pi }{6}} \right) + \sin \dfrac{\pi }{4}\cos \left( {x + \dfrac{\pi }{6}} \right)} \right]\]
Apply the formula \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\].
\[ = \sqrt 2 \sin \left( {x + \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right)\]
We know that, the maxima value of \[\sin \theta \] is 1.
So, \[\sin \left( {x + \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right) = 1\]
We know \[\sin \theta = 1\] at \[\theta = \dfrac{\pi }{2}\] when \[\theta \in \left[ {0,\dfrac{\pi }{2}} \right]\].
Now putting \[\sin \dfrac{\pi }{2}\] in place 1.
\[\sin \left( {x + \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{2}\]
Apply the formula \[\sin \alpha = \sin \beta \Rightarrow \alpha = \beta \]
\[ \Rightarrow \left( {x + \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right) = \dfrac{\pi }{2}\]
\[ \Rightarrow x = \dfrac{\pi }{2} - \dfrac{\pi }{6} - \dfrac{\pi }{4}\]
\[ \Rightarrow x = \dfrac{{6\pi - 2\pi - 3\pi }}{{12}}\]
\[ \Rightarrow x = \dfrac{\pi }{{12}}\]
The expression \[\sin \left( {x + \dfrac{\pi }{6}} \right) + \cos \left( {x + \dfrac{\pi }{6}} \right)\] has maxima at \[x = \dfrac{\pi }{12}\].
Hence option A is the correct.

Note: Students often get confused with the formulas \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\] and \[\sin \left( {a + b} \right) = \sin a\sin b - \cos a\cos b\]. But correct formulas are \[\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b\] and \[\cos \left( {a + b} \right) = \sin a\sin b - \cos a\cos b\].