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.