Question

The most general values of x for which \[\sin x + \cos x = {\min _{a \in \mathbb{R}}}\{ 1,{a^2} - 4a + 6\} \] are given by
a. \[2n\pi ;n \in \mathbb{Z}\]
b. \[2n\pi + \dfrac{\pi }{2};n \in \mathbb{Z}\]
c. \[n\pi + {( - 1)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{4};n \in \mathbb{Z}\]
d. \[2n\pi - \dfrac{\pi }{2};n \in \mathbb{Z}\]

Answer 1

Hint: To start with we can use the fact that what is the minimum value of the given polynomial, \[{a^2} - 4a + 6\]. If the value is less than 1, then we can proceed by using that value for \[\sin x + \cos x\] to find the general value for x.

Complete step-by-step answer:
Given \[\sin x + \cos x = {\min _{a \in \mathbb{R}}}\{ 1,{a^2} - 4a + 6\} \]
We have now,
 \[{a^2} - 4a + 6\]
On splitting last term, we get
 \[ = {a^2} - 4a + 4 + 2\]
Using \[{a^2} - 2ab + {b^2} = {(a - b)^2}\] , we get,
 \[ = {(a - 2)^2} + 2\]
So, minimum, \[{a^2} - 4a + 6\]
 \[ = {(0)^2} + 2\]
On simplification we get,
 \[ = 0 + 2\]\[ = 2\]
So, now, \[\sin x + \cos x = \min (1,{a^2} - 4a + 6)\]
So, the minimum value is, \[\sin x + \cos x = 1\]
Then, \[\sin x + \cos x = 1\]
Multiplying by \[\dfrac{1}{{\sqrt 2 }}\] , we get,
 \[ \Rightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }}\]
Using, \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]and \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\] , we get,
 \[ \Rightarrow \sin \dfrac{\pi }{4}\sin x + \cos \dfrac{\pi }{4}\cos x = \cos \dfrac{\pi }{4}\]
Now using \[\cos (a - b) = \cos a\cos b + \sin a\sin b\] , we get,
 \[ \Rightarrow \cos (x - \dfrac{\pi }{4}) = \cos \dfrac{\pi }{4}\]
Using, \[x = 2n\pi \pm y\] for \[\cos x = \cos y\], we get,
 \[ \Rightarrow x - \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4}\]
On simplification we get,
 \[ \Rightarrow x = 2n\pi \pm \dfrac{\pi }{4} + \dfrac{\pi }{4}\]
Then, \[x = 2n\pi + \dfrac{\pi }{2}\] or \[x = 2n\pi \]
Also, As, \[\sin x + \cos x = 1\]
Multiplying by \[\dfrac{1}{{\sqrt 2 }}\] , we get,
 \[ \Rightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }}\]
Using, \[\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]and \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\], we get,
 \[ \Rightarrow \cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x = \sin \dfrac{\pi }{4}\]
Using \[\sin (a + b) = \sin a\cos b + \cos a\sin b\] , we get,
 \[ \Rightarrow \sin (x + \dfrac{\pi }{4}) = \sin \dfrac{\pi }{4}\]
Using, \[x = n\pi + {( - 1)^n}y\]for \[\sin x = \sin y\]
 \[ \Rightarrow x + \dfrac{\pi }{4} = n\pi + {( - 1)^n}\dfrac{\pi }{4}\]
On simplification we get,
 \[ \Rightarrow x = n\pi + {( - 1)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4}\] where \[n \in \mathbb{Z}\]
We have the general value as, \[x = n\pi + {( - 1)^n}\dfrac{\pi }{4} - \dfrac{\pi }{4}\] where \[n \in \mathbb{Z}\] which is option c.

Note: We have the general value of \[\sin x = \sin y\]as, \[x = n\pi + {( - 1)^n}y\] and also \[\cos x = \cos y\] as, \[x = 2n\pi \pm y\] . Then we get the values of x as, the value of n goes on starting from, \[n = 1,2,3\],……. And so on.