Solveeit Logo
Login

Question

Question: ABC is a triangle, then find \[{a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\c...

ABC is a triangle, then find a2(cos⁡2B−cos⁡2C)+b2(cos⁡2C−cos⁡2A)+c2(cos⁡2A−cos⁡2B){a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right).
A. 0
B. 1
C. a2+b2+c2{a^2} + {b^2} + {c^2}
D. 2(a2+b2+c2)2\left( {{a^2} + {b^2} + {c^2}} \right)

Explanation

Solution

First we will convert all cos⁡\cos function into sin⁡\sin function using trigonometry identity. Then using the sine law, we will find the value of sin⁡A\sin A, sin⁡B\sin B and sin⁡C\sin C. After that we will substitute the value of sin⁡A\sin A, sin⁡B\sin B and sin⁡C\sin C in the given expression.

Formula used:
Sine law
sin⁡Aa=sin⁡Bb=sin⁡Cc\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}
Trigonometry identity
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1

Complete step by step solution:
Given expression is a2(cos⁡2B−cos⁡2C)+b2(cos⁡2C−cos⁡2A)+c2(cos⁡2A−cos⁡2B){a^2}\left( {{{\cos }^2}B - {{\cos }^2}C} \right) + {b^2}\left( {{{\cos }^2}C - {{\cos }^2}A} \right) + {c^2}\left( {{{\cos }^2}A - {{\cos }^2}B} \right)
Now applying the trigonometry identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
=a2(1−sin⁡2B−1+sin⁡2C)+b2(1−sin⁡2C−1+sin⁡2A)+c2(1−sin⁡2A−1+sin⁡2B)= {a^2}\left( {1 - {{\sin }^2}B - 1 + {{\sin }^2}C} \right) + {b^2}\left( {1 - {{\sin }^2}C - 1 + {{\sin }^2}A} \right) + {c^2}\left( {1 - {{\sin }^2}A - 1 + {{\sin }^2}B} \right)
=a2(sin⁡2C−sin⁡2B)+b2(sin⁡2A−sin⁡2C)+c2(sin⁡2B−sin⁡2A)= {a^2}\left( {{{\sin }^2}C - {{\sin }^2}B} \right) + {b^2}\left( {{{\sin }^2}A - {{\sin }^2}C} \right) + {c^2}\left( {{{\sin }^2}B - {{\sin }^2}A} \right) …..(i)
We know that,
sin⁡Aa=sin⁡Bb=sin⁡Cc=k(say)\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c} = k\left( {{\rm{say}}} \right)
sin⁡A=ak\sin A = ak, sin⁡B=bk\sin B = bk, and sin⁡C=ck\sin C = ck
Substitute sin⁡A=ak\sin A = ak, sin⁡B=bk\sin B = bk, and sin⁡C=ck\sin C = ck in the expression (i)
=a2(c2k2−b2k2)+b2(a2k2−c2k2)+c2(b2k2−a2k2)= {a^2}\left( {{c^2}{k^2} - {b^2}{k^2}} \right) + {b^2}\left( {{a^2}{k^2} - {c^2}{k^2}} \right) + {c^2}\left( {{b^2}{k^2} - {a^2}{k^2}} \right)
Simplify the above expression
=a2c2k2−a2b2k2+b2a2k2−b2c2k2+c2b2k2−c2a2k2= {a^2}{c^2}{k^2} - {a^2}{b^2}{k^2} + {b^2}{a^2}{k^2} - {b^2}{c^2}{k^2} + {c^2}{b^2}{k^2} - {c^2}{a^2}{k^2}
Cancel out the opposite term
=0= 0
Hence option A is the correct option.

Note: Students often make a common mistake to solve the given question. They apply cosine law to solve the given question. But the given question should be solved by using trigonometry identity and the sine law.