Question

If a,b, and A are given in a triangle and $c_1,c_2$ are the possibles values of the third side,then $c_1^2+c_2^2-2c_1{c}_2\cos 2A$=is equal to

• A. $4a^2\sin 2A$
• B. $4a^2{\sin }^{2}2A$
• C. $4a^2\cos 2A$
• D. $4a^2{\cos }^{2}A$

Answer 1

The correct option is D $4a^2{\cos }^{2}A$

Since $a,b,c$ are given, so using cosine rule,

$\cos A=\frac{b^2+c^2-a^2}{2bc}\Rightarrow c^2-\left(2b\cos A\right)c+(b^2-a^2)=0$

$\Rightarrow c_1+c_2=2b\cos A,c_1{c}_2=b^2-a^2$ .........(i)

Thus $c_1^2+c_2^2-2c_1{c}_2\cos 2A=(c_1+c_2{)}^2-2c_1{c}_2-2c_1{c}_2\cos 2A$

$=4b^2{\cos }^{2}A-2(1+\cos 2A)(b^2-a^2)=4b^2{\cos }^{2}A-2.2{\cos }^{2}A(b^2-a^2)=4a^2{\cos }^{2}A$