Question

If $ x + y = z $ then find the value of $ {\cos ^2}x + {\cos ^2}y + {\cos ^2}z - 2\cos x\cos y\cos z $ .
(A) $ {\cos ^2}z $
(B) $ {\sin ^2}z $
(C) $ 0 $
(D) $ 1 $

Answer 1

Hint: In this problem, first we will use the trigonometric formula which is given by $ {\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2} $ . Then, we will use another trigonometric formula which is given by $ 2\cos \alpha \cos \beta = \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right) $ . After simplification and using given information $ x + y = z $ , we will use one more trigonometric formula which is given by $ \cos \alpha + \cos \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) $ to find required answer.

Complete step-by-step answer:
In this problem, it is given that $ x + y = z $ and we have to find the value of $ {\cos ^2}x + {\cos ^2}y + {\cos ^2}z - 2\cos x\cos y\cos z \cdots \cdots \left( 1 \right) $
We know that $ {\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2} $ . Use this formula in the first and second term of expression $ \left( 1 \right) $ . So, it can be written as
$\left( {\dfrac{{1 + \cos 2x}}{2}} \right) + \left( {\dfrac{{1 + \cos 2y}}{2}} \right) + \left( {{{\cos}^2}z} \right) - \left( {2\cos x\cos y} \right)\cos z \cdots \cdots \left( 2 \right) $
Also we know that
$ 2\cos \alpha \cos \beta = \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right) $ .
Use this formula in the bracket of the last term of expression $ \left( 2 \right) $ and simplify other terms. So, it can be written as
 $
\Rightarrow \left( {\dfrac{1}{2} + \dfrac{{\cos 2x}}{2}} \right) + \left( {\dfrac{1}{2} + \dfrac{{\cos 2y}}{2}} \right) + \left( {{{\cos }^2}z} \right) - \left[ {\cos \left( {x + y} \right) + \cos \left( {x - y} \right)} \right]\cos z \;
   = \left( {\dfrac{1}{2} + \dfrac{1}{2}} \right) + \dfrac{1}{2}\left( {\cos 2x + \cos 2y} \right) + \left( {{{\cos }^2}z} \right) - \left[ {\cos \left( {x + y} \right)\cos z + \cos \left( {x - y} \right)\cos z} \right] \\
   = 1 + \dfrac{1}{2}\left( {\cos 2x + \cos 2y} \right) + \left( {{{\cos }^2}z} \right) - \cos \left( {x + y} \right)\cos z - \cos \left( {x - y} \right)\cos z \cdots \cdots \left( 3 \right) \;
  $
Also we know that
$ \cos \alpha + \cos \beta = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) $ .
Use this formula in the second term of expression $ \left( 3 \right) $ and then substitute $ x + y = z $ . So, it can be written as
 $
\Rightarrow 1 + \dfrac{1}{2}\left[ {2\cos \left( {\dfrac{{2x + 2y}}{2}} \right)\cos \left( {\dfrac{{2x - 2y}}{2}} \right)} \right] + \left( {{{\cos }^2}z} \right) - \cos z\cos z - \cos \left( {x - y} \right)\cos z \\
 = 1 + \cos \left( {x + y} \right)\cos \left( {x - y} \right) + {\cos ^2}z - {\cos ^2}z - \cos \left( {x - y} \right)\cos z \\
 = 1 + \cos z\cos \left( {x - y} \right) + {\cos ^2}z - {\cos ^2}z - \cos z\cos \left( {x - y} \right) \cdots \cdots \left( 4 \right) \;
  $
By cancelling equal terms with opposite sign from expression $ \left( 4 \right) $ , we get
 $ {\cos ^2}x + {\cos ^2}y + {\cos ^2}z - 2\cos x\cos y\cos z = 1 $
So, the correct answer is “Option D”.

Note: In this type of problems, trigonometric identities and formulas are very useful to find the required answer. Here we used the formula $ {\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2} $ at the first step of solution but we can start with the formula $ 2\cos \alpha \cos \beta = \cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right) $ . Also we can use the trigonometric formula $ \cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y $ and Pythagorean identity $ {\sin ^2}x + {\cos ^2}x = 1 $ in the given problem.