Question

Maximum value of the function $\sin \left( {\theta + \dfrac{\pi }{6}} \right) + \cos \left( {\theta + \dfrac{\pi }{6}} \right)$ in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ is attained at

Answer 1

Hint: For solving this particular solution , firstly multiply and divide the given expression by $\sqrt 2 $ . then substitute $\dfrac{1}{{\sqrt 2 }} = \cos \left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)$ . Then by using the trigonometry identity which says, $\sin (A + B) = \sin A\cos B + \cos A\sin B$ we will get the required result. Lastly simplify the equation.

Complete step by step solution:
we have to find the maximum value of the function $\sin \left( {\theta + \dfrac{\pi }{6}} \right) + \cos \left( {\theta + \dfrac{\pi }{6}} \right)$ in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ .
let us take the function and try to simplify it ,
$ \Rightarrow \sin \left( {\theta + \dfrac{\pi }{6}} \right) + \cos \left( {\theta + \dfrac{\pi }{6}} \right)$ (given)
Multiply and divide the above equation by $\sqrt 2 $ . we will get ,
$ \Rightarrow \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin \left( {\theta + \dfrac{\pi }{6}} \right) + \dfrac{1}{{\sqrt 2 }}\cos \left( {\theta + \dfrac{\pi }{6}} \right)} \right)$
As we know that $\dfrac{1}{{\sqrt 2 }} = \cos \left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right)$ . Now put this in above equation,
\[
   \Rightarrow \sqrt 2 \left( {\cos \left( {\dfrac{\pi }{4}} \right)\sin \left( {\theta + \dfrac{\pi }{6}} \right) + \sin \left( {\dfrac{\pi }{4}} \right)\cos \left( {\theta + \dfrac{\pi }{6}} \right)} \right) \\
   \Rightarrow \sqrt 2 \left( {\sin \left( {\theta + \dfrac{\pi }{6} + \dfrac{\pi }{4}} \right)} \right) \\
 \]
By using the trigonometry identity which says, $\sin (A + B) = \sin A\cos B + \cos A\sin B$ .
We also know that the maximum value of sine function is one. Sine is one at $\dfrac{\pi }{2}$ in the interval of $\left[ {0,\dfrac{\pi }{2}} \right]$ therefore , equate as follow,
$ \Rightarrow \theta + \dfrac{\pi }{6} + \dfrac{\pi }{4} = \dfrac{\pi }{2}$
$ \Rightarrow \theta + \dfrac{\pi }{6} + \dfrac{\pi }{4} = \dfrac{\pi }{2}$
$ \Rightarrow \theta = \dfrac{\pi }{2} - \dfrac{\pi }{6} - \dfrac{\pi }{4}$
$ = \dfrac{\pi }{{12}}$
Therefore , the required solution is $\dfrac{\pi }{{12}}$ .

Note:
The maximum and minimum value of cosine and sine with odd power is plus one and minus one.
The maximum and minimum value of cosine and sine with even power is plus one and zero.
The maximum and minimum value of tangent and cotangent with odd power is plus infinity and minus infinity .
The maximum and minimum value of tangent and cotangent with even power is plus infinity and zero.
The maximum and minimum value of secant and cosecant with odd power is plus infinity and minus infinity.
The maximum and minimum value of tangent and cotangent with even power is plus infinity and one.