Question

In a triangle ABC, if \[{a^4} + {b^4} + {c^4} = 2{a^2}{b^2} + {b^2}{c^2} + 2{c^2}{a^2}\], then find the value of $\sin A$.

Answer 1

Hint:First, move all the variables on one side of the equation and add \[3{b^2}{c^2}\] on both sides of the equation. Then, use the formula \[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\] to convert the equation into cosine rule of $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$. Then find the value of angle A and use it to find the value of $\sin A.$

Formula used:$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
\[{\left( {p + q + r} \right)^2} = {p^2} + {q^2} + {r^2} + 2pq + 2qr + 2rp\]

Complete step-by-step answer:
Given equation \[{a^4} + {b^4} + {c^4} = 2{a^2}{b^2} + {b^2}{c^2} + 2{c^2}{a^2}\]
We have to find the value of sin A.
Shift all the variables to one side of the equation,
\[{a^4} + {b^4} + {c^4} - 2{a^2}{b^2} - {b^2}{c^2} - 2{c^2}{a^2} = 0\]
Add \[3{b^2}{c^2}\] to both sides of the equation,
\[{a^4} + {b^4} + {c^4} - 2{a^2}{b^2} - {b^2}{c^2} - 2{c^2}{a^2} + 3{b^2}{c^2} = 3{b^2}{c^2}\]
Add the like terms to simplify the equation in left side of the equation,
\[{a^4} + {b^4} + {c^4} - 2{a^2}{b^2} + 2{b^2}{c^2} - 2{c^2}{a^2} = 3{b^2}{c^2}\]
As we know that \[{\left( {p + q + r} \right)^2} = {p^2} + {q^2} + {r^2} + 2pq + 2qr + 2rp\]. Then make the L.H.S. of the equation in this form,
\[{\left( {{a^2}} \right)^2} + {\left( {{b^2}} \right)^2} + {\left( {{c^2}} \right)^2} - 2\left( {{a^2}} \right)\left( {{b^2}} \right) + 2\left( {{b^2}} \right)\left( {{c^2}} \right) - 2\left( {{c^2}} \right)\left( {{a^2}} \right) = 3{b^2}{c^2}\]
Use negative sign with the terms as needed,
\[{\left( { - {a^2}} \right)^2} + {\left( {{b^2}} \right)^2} + {\left( {{c^2}} \right)^2} + 2\left( { - {a^2}} \right){b^2} + 2{b^2}{c^2} + 2\left( {{c^2}} \right)\left( { - {a^2}} \right) = 3{b^2}{c^2}\]
Replace \[{\left( { - {a^2}} \right)^2} + {\left( {{b^2}} \right)^2} + {\left( {{c^2}} \right)^2} + 2\left( { - {a^2}} \right){b^2} + 2{b^2}{c^2} + 2\left( {{c^2}} \right)\left( { - {a^2}} \right)\] by \[{\left( {{b^2} + {c^2} - {a^2}} \right)^2}\] to simplify the equation,
\[{\left( {{b^2} + {c^2} - {a^2}} \right)^2} = 3{b^2}{c^2}\]
Take square root on both sides of the equation,
\[\sqrt {{{\left( {{b^2} + {c^2} - {a^2}} \right)}^2}} = \sqrt {3{b^2}{c^2}} \]
Cancel out squares with the square root, we get,
\[{b^2} + {c^2} - {a^2} = \sqrt 3 bc\]
Now, divide both sides of the equation by $2bc$ we get,
\[\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \dfrac{{\sqrt 3 bc}}{{2bc}}\]
Cancel out the common factors,
$\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \dfrac{{\sqrt 3 }}{2}$
As we know that the cosine rule of $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$. Replace the term by $\cos A$,
\[\cos A = \dfrac{{\sqrt 3 }}{2}\]
Since, $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$. Replace $\dfrac{{\sqrt 3 }}{2}$ by $\cos \dfrac{\pi }{6}$,
$\cos A = \cos \dfrac{\pi }{6}$
Thus, $A = \dfrac{\pi }{6}$
Now, find the value of $\sin A$,
$\sin A = \sin \dfrac{\pi }{6}$
We know from trigonometric standard angles table $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
So, $\sin A = \dfrac{1}{2}$

So, the correct answer is “Option D”.

Note:This type of problem can be solved by performing the cosine rule of $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$. Students should be taken care of when adding the like terms, changing the signs. Students must recall the algebraic identity,trigonometry formula and trigonometric standard angles for solving these types of problems.