Question

If \[0 < x,y < \pi \] and \[\cos x + \cos y - \cos \left( {x + y} \right) = \dfrac{3}{2}\]then find the value of \[\sin x + \cos y\].
A. \[\dfrac{{1 + \sqrt 3 }}{2}\]
B. \[\dfrac{{1 - \sqrt 3 }}{2}\]
C. \[\dfrac{{\sqrt 3 }}{2}\]
D. \[\dfrac{1}{2}\]

Answer 1

Hint: In this question, for determining the value of \[\sin x + \cos y\], we need to simplify the expression \[\cos x + \cos y - \cos \left( {x + y} \right) = \dfrac{3}{2}\]. For this, we need to use trigonometric identities such as \[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] and the half angle formula such as \[\cos A = 2{\cos ^2}\left( {A/2} \right) - 1\]


Complete step by step answer: We know that \[\cos x + \cos y - \cos \left( {x + y} \right) = \dfrac{3}{2}\].
Let us simplify this expression to get the desired result.
Now, consider
\[\cos x + \cos y - \cos \left( {x + y} \right) = \dfrac{3}{2}\]
Let us us the formula such as \[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
\[
   \Rightarrow 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) - \left[ {2{{\cos }^2}\left( {\dfrac{{x + y}}{2}} \right) - 1} \right] = \dfrac{3}{2} \\
   \Rightarrow 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) - 2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) = \dfrac{1}{2} \\
 \]
Multiply by \[2\] to both sides.
Thus, we get
\[
   \Rightarrow 4\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) - 4{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) = 1 = {\cos ^2}\left( {\dfrac{{x - y}}{2}} \right) + {\sin ^2}\left( {\dfrac{{x - y}}{2}} \right) \\
   \Rightarrow 4{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) + {\cos ^2}\left( {\dfrac{{x - y}}{2}} \right) - 4\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) + {\sin ^2}\left( {\dfrac{{x - y}}{2}} \right) = 0 \\
   \Rightarrow {\left( {\cos \left( {\dfrac{{x - y}}{2}} \right) - 2\cos \left( {\dfrac{{x + y}}{2}} \right)} \right)^2} + {\sin ^2}\left( {\dfrac{{x - y}}{2}} \right) = 0 \\
   \Rightarrow \sin \left( {\dfrac{{x - y}}{2}} \right) = 0 \Rightarrow x = y \\
 \]
Also, we get
\[
  \cos \left( {\dfrac{{x - y}}{2}} \right) = 2\cos \left( {\dfrac{{x + y}}{2}} \right) \\
   \Rightarrow \cos x = \dfrac{1}{2} = \cos y \\
 \]
Hence, the value of \[\sin x + \cos y\] can be calculated as
\[
  \sin x + \cos y = \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} \\
   \Rightarrow \sin x + \cos y = \dfrac{{\sqrt 3 + 1}}{2} \\
 \]
So, the value of \[\sin x + \cos y\] is \[\dfrac{{1 + \sqrt 3 }}{2}\] if \[\cos x + \cos y - \cos \left( {x + y} \right) = \dfrac{3}{2}\]

Therefore, the option (A) is correct

Note: Many students make mistakes in solving the calculation part and applying trigonometric identities. This is the only way, through which we can solve the example in the simplest way. Using proper trigonometric identities is necessary for solving trigonometric problems as this makes them simple.