Question

If we have an inequality as \[\sin x+\sin y\ge \cos \alpha \cos x,\forall x\in R\], then \[\sin y+ \cos \alpha \] is equal to,
A. -1
B. 1
C. 0
D. 2

Answer 1

Hint: We will be using the concepts of trigonometric functions to solve the problem. We are going to rearrange the above inequality and then we will find the greatest value of the trigonometric expression. From this, we will get the value of $\sin y\And \cos \alpha $ and after that we will put those values in that expression.

Complete step-by-step solution
Now, we have been given that \[\sin x+\sin y\ge \cos \alpha \cos x,\forall x\in R\]. Since, it is true for all \[x\in R\].
We are going to subtract $\sin x$ on both the sides of the above inequality and we get,
$\begin{align}
  & \Rightarrow \sin x-\sin x+\sin y\ge \cos \alpha \cos x-\sin x \\
 & \Rightarrow \sin y\ge \cos \alpha \cos x-\sin x \\
\end{align}$
If we find the value of $\sin y$ in such a way so that its value is always greater than $\cos \alpha \cos x-\sin x$ then no matter what x can be the above inequality holds always true so we are going to find the maximum value of $\cos \alpha \cos x-\sin x$. For that we are going to multiply and divide the R.H.S of the above inequality by $\sqrt{1+{{\cos }^{2}}\alpha }$ we get,
$\Rightarrow \sin y\ge \sqrt{1+{{\cos }^{2}}\alpha }\left( \dfrac{\cos \alpha \cos x}{\sqrt{1+{{\cos }^{2}}\alpha }}-\dfrac{\sin x}{\sqrt{1+{{\cos }^{2}}\alpha }} \right)$ ……………..(1)
Let us assume $\dfrac{1}{\sqrt{1+{{\cos }^{2}}\alpha }}=\sin \theta $ then we are going to use the property of $\cos \theta \And \sin \theta $ to find the value of $\cos \theta $ which is equal to:
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Substituting the value of $\sin \theta $ in the above equation we get,
$\begin{align}
  & \Rightarrow \cos \theta =\sqrt{1-\dfrac{1}{1+{{\cos }^{2}}\alpha }} \\
 & \Rightarrow \cos \theta =\sqrt{\dfrac{1+{{\cos }^{2}}\alpha -1}{1+{{\cos }^{2}}\alpha }} \\
 & \Rightarrow \cos \theta =\sqrt{\dfrac{{{\cos }^{2}}\alpha }{1+{{\cos }^{2}}\alpha }} \\
\end{align}$
$\Rightarrow \cos \theta =\dfrac{\cos \alpha }{\sqrt{1+{{\cos }^{2}}\alpha }}$
So, substituting the above values of $\dfrac{\cos \alpha }{\sqrt{1+{{\cos }^{2}}\alpha }}\And \dfrac{1}{\sqrt{1+{{\cos }^{2}}\alpha }}$ as $\cos \theta \And \sin \theta $ in (1) we get,
$\Rightarrow \sin y\ge \sqrt{1+{{\cos }^{2}}\alpha }\left( \cos \theta \cos x-\sin \theta \sin x \right)$
We know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ so using this trigonometric relation in the above problem we get,
$\Rightarrow \sin y\ge \sqrt{1+{{\cos }^{2}}\alpha }\left( \cos \left( \theta +x \right) \right)$
The maximum value of the R.H.S is $\sqrt{1+{{\cos }^{2}}\alpha }$ then the above inequality will look like:
$\Rightarrow \sin y\ge \sqrt{1+{{\cos }^{2}}\alpha }$
Now, the maximum value which $\sin y$ can take is 1 and we are taking the equality sign so $\sqrt{1+{{\cos }^{2}}\alpha }$ also equals 1.
And $\sqrt{1+{{\cos }^{2}}\alpha }=1$ when $\cos \alpha =0$. Hence, we got the value of $\sin y=1\And \cos \alpha =0$.
Now, in the above problem we are asked to find the value of the expression $\sin y+\cos \alpha $. Substituting the values of $\sin y=1\And \cos \alpha =0$ in this expression we get,
$\begin{align}
  & \Rightarrow \sin y+\cos \alpha \\
 & =1+0 \\
 & =1 \\
\end{align}$
Hence, the correct option is (b).

Note: In the above solution, we have found the maximum value of $\cos \alpha \cos x-\sin x$ by using the formula for the greatest value of $a\cos x\pm b\sin x$ which is equal to $\sqrt{{{a}^{2}}+{{b}^{2}}}$. Now, on comparing “a” and “b” values with $\cos \alpha \And -1$ respectively then the maximum value of $\cos \alpha \cos x-\sin x$ is equal to:
$\sqrt{1+{{\cos }^{2}}\alpha }$