Question

If $A,B,C$ are the angles of a triangle, then the value of ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A.\cos B.\cos c =$?

Answer 1

In order to solve this question first, we assume a variable equal to the asked expression. Then we multiply and divide by 2 into the first three factors. And then convert those in cos double angle. Then add the constant part and use the sum of the two cos functions and express those in terms of the third angle. And then take the common and remaining last angle in terms of the first two angles then again use the sum of cos function. And then make some calculations and get some calculations.

Complete step by step answer:
We have given that $A,B,C$ are the angles of a triangle.
To find,
The value of the ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A.\cos B.\cos c =$ expression
Let the obtained value from the asked expression is $x$
$x = {\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A.\cos B.\cos c$ (given)
On multiplying and dividing by 2 in the first three terms.
$\Rightarrow \dfrac{1}{2}\left( {2{{\sin }^2}A + 2{{\sin }^2}B + 2{{\sin }^2}C} \right) - 2\cos A.\cos B.\cos c$
Now using the formula of double angle $2{\sin ^2}\theta = 1 - \cos 2\theta$
$\Rightarrow \dfrac{1}{2}\left( {1 - \cos 2A + 1 - \cos 2B + 1 - \cos 2c} \right) - 2\cos A.\cos B.\cos c$
On rearranging the expression
$\Rightarrow \dfrac{1}{2}\left( {3 - \left( {\cos 2A + \cos 2B + \cos 2c} \right)} \right) - 2\cos A.\cos B.\cos c$
Now using the formula $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos 2\theta = 2{\cos ^2}\theta - 1$
$\Rightarrow \dfrac{1}{2}\left( {3 - 2\cos \left( {A + B} \right)\cos \left( {A - B} \right) - 2{{\cos }^2}c + 1} \right) - 2\cos A.\cos B.\cos c$
Now adding the constant term and we know that sum of all angle of triangle is $\pi$ so we write $A + B = \pi - C$
$\Rightarrow \dfrac{1}{2}\left( {4 - 2\cos \left( {\pi - C} \right)\cos \left( {A - B} \right) - 2{{\cos }^2}c} \right) - 2\cos A.\cos B.\cos c$
We know that $\cos \left( {\pi - C} \right) = - \cos C$
$\Rightarrow \dfrac{1}{2}\left( {4 + 2\cos C\cos \left( {A - B} \right) - 2{{\cos }^2}c} \right) - 2\cos A.\cos B.\cos c$
Now on multiplying and expanding.
$\Rightarrow 2 + \cos C\cos \left( {A - B} \right) - {\cos ^2}C - 2\cos A.\cos B.\cos C$
Taking $\cos C$ common from the first two terms and writing $\pi - \left( {A + B} \right)$.
$\Rightarrow 2 + \cos C\left( {\cos \left( {A - B} \right) - \cos \left( {\pi - \left( {A + B} \right)} \right)} \right) - 2\cos A.\cos B.\cos C$
We know that $\cos \left( {\pi - C} \right) = - \cos C$
$\Rightarrow 2 + \cos C\left( {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right) - 2\cos A.\cos B.\cos C$
Now using the formula $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
$\Rightarrow 2 + 2\cos C\cos A\cos B - 2\cos A.\cos B.\cos C$
On simplifying we get-
$\Rightarrow 2$
If $A,B,C$ are the angles of a triangle, then the value of ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A.\cos B.\cos c = 2$

Note: To solve this type of question students must know all the formulas of trigonometry and conditions of the triangle. There are many places where students get confused or stuck in between so practice more questions which will increase the solving speed and accuracy.