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Question: If in a triangle ABC, \(\theta \) is the angle determined by \(\cos \theta = \left( {\dfrac{{a - b}}...

If in a triangle ABC, θ\theta is the angle determined by cos⁡θ=(a−bc)\cos \theta = \left( {\dfrac{{a - b}}{c}} \right), then which of the following is correct?
a.(a+b)sin⁡θ2ab=cos⁡A−B2\dfrac{{\left( {a + b} \right)\sin \theta }}{{2\sqrt {ab} }} = \cos \dfrac{{A - B}}{2}
b.(a+b)sin⁡θ2ab=cos⁡A+B2\dfrac{{\left( {a + b} \right)\sin \theta }}{{2\sqrt {ab} }} = \cos \dfrac{{A + B}}{2}
c.csin⁡θ2ab=cos⁡A−B2\dfrac{{c\sin \theta }}{{2\sqrt {ab} }} = \cos \dfrac{{A - B}}{2}
d.csin⁡θ2ab=cos⁡A+B2\dfrac{{c\sin \theta }}{{2\sqrt {ab} }} = \cos \dfrac{{A + B}}{2}

Explanation

Solution

To check the availability of the options, we will check every option and examine if their left hand side is equal to right hand side or not. If yes, then we will say that option is correct and it satisfies the question.

Complete step-by-step answer:
We are given cos⁡θ=(a−bc)\cos \theta = \left( {\dfrac{{a - b}}{c}} \right)
Also, we know the trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Therefore, we can say that sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta
⇒\Rightarrow sin⁡θ=1−cos⁡2θ=1−(a−bc)2=1cc2−(a−b)2\sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - {{\left( {\dfrac{{a - b}}{c}} \right)}^2}} = \dfrac{1}{c}\sqrt {{c^2} - {{\left( {a - b} \right)}^2}}
For option (a), considering L.H.S. and substituting the value of sin⁡θ\sin \theta , we get
(a+b)sin⁡θ2ab=(a+b)c2−(a−b)22cab=(a+b)a2+b2−c2+2ab2cab\dfrac{{\left( {a + b} \right)\sin \theta }}{{2\sqrt {ab} }} = \dfrac{{\left( {a + b} \right)\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} }}{{2c\sqrt {ab} }} = \dfrac{{\left( {a + b} \right)\sqrt {{a^2} + {b^2} - {c^2} + 2ab} }}{{2c\sqrt {ab} }}
⇒(a+b)2ab(1−cos⁡c)2cab=(a+b)c22sin⁡(c2)\Rightarrow \dfrac{{\left( {a + b} \right)\sqrt {2ab\left( {1 - \cos c} \right)} }}{{2c\sqrt {ab} }} = \dfrac{{\left( {a + b} \right)}}{{c\sqrt 2 }}\sqrt 2 \sin \left( {\dfrac{c}{2}} \right)

⇒sin⁡A+sin⁡Bsin⁡Csin⁡(c2)=2sin⁡(A+B2)cos⁡(A−B2)2sin⁡(C2)cos⁡(C2)sin⁡(C2)=cos⁡(A−B2) \Rightarrow \dfrac{{\sin A + \sin B}}{{\sin C}}\sin \left( {\dfrac{c}{2}} \right) = \dfrac{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}{{2\sin \left( {\dfrac{C}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)}}\sin \left( {\dfrac{C}{2}} \right) = \cos \left( {\dfrac{{A - B}}{2}} \right)= R.H.S
Hence, option (a) is correct.
For option (b), it can’t be true since option(a) is true. This is because it is mentioned in the option (b) that its left hand side is equal to cos⁡(A−B2)\cos \left( {\dfrac{{A - B}}{2}} \right)but that can’t be possible because we have proved that it is equal to cos⁡(A+B2)\cos \left( {\dfrac{{A + B}}{2}} \right).
For option (c), considering L.H.S. of the equation,
csin⁡θ2ab=cc2−(a−b)22ab=c2ab(1−cos⁡c)2cab\dfrac{{c\sin \theta }}{{2\sqrt {ab} }} = \dfrac{{c\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} }}{{2\sqrt {ab} }} = \dfrac{{c\sqrt {2ab\left( {1 - \cos c} \right)} }}{{2c\sqrt {ab} }}
⇒sin⁡c2=cos⁡A+B2\Rightarrow \sin \dfrac{c}{2} = \cos \dfrac{{A + B}}{2} ≠\neR.H.S.
Hence, option (c) is not correct.
For option (d), it is true as we just proved it in option (c).
Hence option C and D are true.

Note: When we come across such a problem, then we check each option as it may have more than 2 correct options. And you may get confused in trigonometric identities i.e., which identity should be used. Trigonometric identities are defined as: the equalities in mathematics that involve trigonometric functions and they hold true for each and every value of their variables given that both sides of the given equality are defined.