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# If $\left( {1 - \cos A} \right)\left( {1 - \cos B} \right)\left( {1 - \cos C} \right) = \sin A\sin B\sin C,$ then find the value of $\left( {1 + \cos A} \right)\left( {1 + \cos B} \right)\left( {1 + \cos C} \right) =$a) $\cos A\cos B\cos C$ b) $\sin A\sin B\sin C$ c) $- \cos A\cos B\cos C$ d) $- \sin A\sin B\sin C$

Last updated date: 25th Jun 2024
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Hint: Here we will use the basic trigonometric identities for equating both the sides of = sign, so that we can reach to the solution. To solve this problem, we need to remember the formulas of trigonometry & apply those attentively.

Given- $\left( {1 - \cos A} \right)\left( {1 - \cos B} \right)\left( {1 - \cos C} \right) = \sin A\sin B\sin C$
We have to find here $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$.
We know some formulas from sub multiple angles of trigonometry & generally we apply these formulas in case half of the angles are involved in the question along with different trigonometric functions -
$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}$
From the given condition, we have
$\left( {1 - \cos A} \right)\left( {1 - \cos B} \right)\left( {1 - \cos C} \right) = \sin A\sin B\sin C$
$2{\sin ^2}\dfrac{A}{2}.2{\sin ^2}\dfrac{B}{2}.2{\sin ^2}\dfrac{C}{2} = \sin A\sin B\sin C$
[ applying formula $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$]
$\Rightarrow 8{\sin ^2}\dfrac{A}{2}{\sin ^2}\dfrac{B}{2}{\sin ^2}\dfrac{C}{2} = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}.2\sin \dfrac{B}{2}\cos \dfrac{B}{2}.2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$$\Rightarrow 8{\sin ^2}\dfrac{A}{2}{\sin ^2}\dfrac{B}{2}{\sin ^2}\dfrac{C}{2} = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}.2\sin \dfrac{B}{2}\cos \dfrac{B}{2}.2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$
[ multiplying all the numerical multiples in one side & $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ on the other ]

$\Rightarrow 8{\sin ^2}\dfrac{A}{2}{\sin ^2}\dfrac{B}{2}{\sin ^2}\dfrac{C}{2} = 8\sin \dfrac{A}{2}\cos \dfrac{A}{2}.\sin \dfrac{B}{2}\cos \dfrac{B}{2}.\sin \dfrac{C}{2}\cos \dfrac{C}{2}$
[ dividing $8\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$from both sides.]
$\Rightarrow \sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2} = \cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$
Squaring both sides, we have
${\sin ^2}\dfrac{A}{2}{\sin ^2}\dfrac{B}{2}{\sin ^2}\dfrac{C}{2} = {\cos ^2}\dfrac{A}{2}{\cos ^2}\dfrac{B}{2}{\cos ^2}\dfrac{C}{2}$…………. (1)
Now,
$\left( {1 + \cos A} \right)\left( {1 + \cos B} \right)\left( {1 + \cos C} \right)$
$= 2{\cos ^2}\dfrac{A}{2}.2{\cos ^2}\dfrac{B}{2}.2{\cos ^2}\dfrac{C}{2}$[ Getting all the numerical multiples multiplied]
$= 8{\cos ^2}\dfrac{A}{2}{\cos ^2}\dfrac{B}{2}{\cos ^2}\dfrac{C}{2}$
$= 8{\sin ^2}\dfrac{A}{2}{\sin ^2}\dfrac{B}{2}{\sin ^2}\dfrac{C}{2}$ [Putting the value of ${\cos ^2}\dfrac{A}{2}{\cos ^2}\dfrac{B}{2}{\cos ^2}\dfrac{C}{2}$ from eq. (1)]
$= 2{\sin ^2}\dfrac{A}{2}.2{\sin ^2}\dfrac{B}{2}.2{\sin ^2}\dfrac{C}{2}$
[ To get in such form so that having formula $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$can be applied ]
$= \left( {1 - \cos A} \right)\left( {1 - \cos B} \right)\left( {1 - \cos C} \right)$
$= \sin A\sin B\sin C$
Hence, the correct option is B.

Note: To solve this we should have concepts & formulas very clear in our mind. In this type of problem, one should take the hint from the given condition in the question and apply it to obtain the solution. Thus, all the trigonometric identities should be remembered. There is a high chance of getting confused while using formulas in different places.
Formulas should be remembered are:
$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}$