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Question: In \(\vartriangle ABC\) , the sides opposite to angles A, B and C are denoted by a, b, c respectivel...

In △ABC\vartriangle ABC , the sides opposite to angles A, B and C are denoted by a, b, c respectively. If 2b2=a2+c22{b^2} = {a^2} + {c^2} , then sin⁡3Bsin⁡B\dfrac{{\sin 3B}}{{\sin B}} is equal to:
A) c2−a22ca\dfrac{{{c^2} - {a^2}}}{{2ca}}
B) c2−a2ca\dfrac{{{c^2} - {a^2}}}{{ca}}
C) (c2−a2ca)2{\left( {\dfrac{{{c^2} - {a^2}}}{{ca}}} \right)^2}
D) (c2−a22ca)2{\left( {\dfrac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}

Explanation

Solution

This problem is based on trigonometric rules where knowing some of the trigonometric formulas will yield the solution. Some of the needed trigonometric formulas are:
sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Also we are in need of cosine rule in a triangle which is applicable for all the three angles of the triangle. As we are in need of the angle B, it is given by,
cos⁡B=a2+c2−b22ac\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}

Complete step-by-step answer:
Step 1: Starting with what we need which is sin⁡3Bsin⁡B\dfrac{{\sin 3B}}{{\sin B}}.
We have the relation sin⁡3B=3sin⁡B−4sin⁡3B\sin 3B = 3\sin B - 4{\sin ^3}B . Thus substituting this in the needed ratio we get,
sin⁡3Bsin⁡B=3sin⁡B−4sin⁡3Bsin⁡B\dfrac{{\sin 3B}}{{\sin B}} = \dfrac{{3\sin B - 4{{\sin }^3}B}}{{\sin B}}
Simplifying we get,
=sin⁡B(3−4sin⁡2B)sin⁡B =3−4sin⁡2B  = \dfrac{{\sin B(3 - 4{{\sin }^2}B)}}{{\sin B}} \\\ = 3 - 4{\sin ^2}B \\\
Step 2: Now we have sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 implies sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta .
Thus we get,
sin⁡3Bsin⁡B=3−4(1−cos⁡2B)\dfrac{{\sin 3B}}{{\sin B}} = 3 - 4(1 - {\cos ^2}B)
=3−4+4cos⁡2B =4cos⁡2B−1  = 3 - 4 + 4{\cos ^2}B \\\ = 4{\cos ^2}B - 1 \\\
Step 3: We have cosine rule with respect to the angle B given by,
cos⁡B=a2+c2−b22ac\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}
Substituting this we get,
sin⁡3Bsin⁡B=4(a2+c2−b22ac)2−1\dfrac{{\sin 3B}}{{\sin B}} = 4{\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)^2} - 1
=4((a2+c2−b2)24a2c2)−1= 4\left( {\dfrac{{{{({a^2} + c{}^2 - {b^2})}^2}}}{{4{a^2}{c^2}}}} \right) - 1
Cancelling 4 and simplifying the above by applying the given relation: 2b2=a2+c22{b^2} = {a^2} + {c^2}, we get
sin⁡3Bsin⁡B=(2b2−b2)2a2c2−1\dfrac{{\sin 3B}}{{\sin B}} = \dfrac{{{{(2{b^2} - {b^2})}^2}}}{{{a^2}{c^2}}} - 1
=b4a2c2−1= \dfrac{{{b^4}}}{{{a^2}{c^2}}} - 1
=b4−a2c2a2c2= \dfrac{{{b^4} - {a^2}{c^2}}}{{{a^2}{c^2}}} … Formula1
Step 4: As we have the relation, 2b2=a2+c22{b^2} = {a^2} + {c^2}
Squaring this we get: b4=(a2+c22)2{b^4} = {\left( {\dfrac{{{a^2} + {c^2}}}{2}} \right)^2}
Substituting the above in Formula 1, we get
sin⁡3Bsin⁡B=a4+c4+2a2c2−4a2c24a2c2\dfrac{{\sin 3B}}{{\sin B}} = \dfrac{{\dfrac{{{a^4} + {c^4} + 2{a^2}{c^2} - 4{a^2}{c^2}}}{4}}}{{{a^2}{c^2}}}
Simplifying the above,
=a4+c4+2a2c2−4a2c24a2c2= \dfrac{{{a^4} + {c^4} + 2{a^2}{c^2} - 4{a^2}{c^2}}}{{4{a^2}{c^2}}}

= \dfrac{{{a^4} + {c^4} - 2{a^2}{c^2}}}{{4{a^2}{c^2}}} \\\ = {\left( {\dfrac{{{c^2} - {a^2}}}{{2ca}}} \right)^2} $$ (Using the relation ${(a - b)^2} = {a^2} + {b^2} - 2ab$) Which is option D. **Option D is correct. $\dfrac{{\sin 3B}}{{\sin B}} = {\left( {\dfrac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}$** **Note:** The steps where errors can occur are the mistakes which happen while applying the trigonometric relations. Proper simplification and application of appropriate trigonometric rules will yield the solution. Some of the basic trigonometric formulas include: $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta $ $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta $ $\sin 2\theta = 2\sin \theta \cos \theta $ $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $ And cosine rules are applicable for all the angles A, B, and C of the triangle given by, $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ $\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$ $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$