Question

If \[\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}\] where \[x\in [0,\pi ]\] then,

a.\[x=\dfrac{\pi }{4},y=1\]

b.\[x=\dfrac{\pi }{4},y=0\]

c.\[x=1,y=1\]

d.\[x=\dfrac{3\pi }{4},y=1\]

Answer 1

Hint: We will first of all use the fact that the minimum value of \[y+\dfrac{1}{y}\] is 2. After this we will apply necessary properties of trigonometric identities such as \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] or any other identities to solve the question completely.

Complete step-by-step answer:
We are given that if \[\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}\]where\[x\in [0,\pi ]\] then we have to find the value of x and y.

To proceed the question, we use the given condition \[\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}\] and try to find the value of x and y.

We know that the minimum value of \[y+\dfrac{1}{y}\] is 2 that is,
\[y+\dfrac{1}{y}\ge 2\]
\[\Rightarrow \sqrt{y+\dfrac{1}{y}}\ge \sqrt{2}\]

Using the obtained expression in the value of \[\sin x+\operatorname{cosx}=\sqrt{y+\dfrac{1}{y}}\] we get,
\[\begin{align}
  & \Rightarrow \sin x+\cos x=\sqrt{y+\dfrac{1}{y}}\ge \sqrt{2} \\
 & \Rightarrow \sin x+\cos x\ge \sqrt{2} \\
\end{align}\]

Squaring both sides of the above expression,
\[\begin{align}
  & \Rightarrow {{\left( \sin x+\cos x \right)}^{2}}\ge 2 \\
 & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x\ge 2 \\
\end{align}\]

We have a trigonometric identity given as,
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]

Using this identity in the above obtained expression we get,
\[1+2\sin x\cos x\ge 2\]

We will again use a trigonometric identity given as,
\[2\sin x\cos x=\sin 2x\]

Using this identity in the above obtained expression we get,
\[\begin{align}
  & 1+\sin 2x\ge 2 \\
 & \Rightarrow \sin 2x\ge 1 \\
\end{align}\]

Now because the maximum value of sin is equal to 1, therefore \[\sin 2x\ge 1\Rightarrow \sin 2x=1\]
\[\begin{align}
  & \Rightarrow \sin 2x=1 \\
 & \Rightarrow 2x=\dfrac{\pi }{2} \\
 & \Rightarrow x=\dfrac{\pi }{4} \\
\end{align}\]

Hence, we got \[x=\dfrac{\pi }{4}\] thus our answer reduces to two options (a) and (b).

Now observing option (b) we get y=0, which is not possible because this y=0 makes the term \[\sqrt{y+\dfrac{1}{y}}\] tends to infinity as we have \[\dfrac{1}{y}\]tends to infinity as y tends to 0.

Hence, we get our answer as \[x=\dfrac{\pi }{4}\] and y=0 which is option (a).

Note: The possibility of error in this question is that when you obtain \[x=\dfrac{\pi }{4}\] as result you may go for opting y=0 as the answer, which is wrong because y=0 will make the expression \[\sqrt{y+\dfrac{1}{y}}\] tends to infinity as we have \[\dfrac{1}{y}\] tends to infinity when y=0.