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Question: Prove that \[\dfrac{{\sin B}}{{\sin A}} = \dfrac{{\sin (2A + B)}}{{\sin A}} - 2\cos (A + B)\]...

Prove that sin⁡Bsin⁡A=sin⁡(2A+B)sin⁡A−2cos⁡(A+B)\dfrac{{\sin B}}{{\sin A}} = \dfrac{{\sin (2A + B)}}{{\sin A}} - 2\cos (A + B)

Explanation

Solution

In proving any statement we first need to expand the given functions, here we have some trigonometric functions so we will try to expand trigonometry identities and try to simplify the function to get the function that is on the other side.
Formula: Formulas that we will be using in this problem:
(1) sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
(2) cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B
(3) sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
(4) cos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2{\sin ^2}A

Complete step by step answer:
It is given that sin⁡Bsin⁡A=sin⁡(2A+B)sin⁡A−2cos⁡(A+B)\dfrac{{\sin B}}{{\sin A}} = \dfrac{{\sin (2A + B)}}{{\sin A}} - 2\cos (A + B) . Let us take the rights side of the given equation and prove it to be the left-hand side.
Let us consider the right-hand side that is sin⁡(2A+B)sin⁡A−2cos⁡(A+B)\dfrac{{\sin (2A + B)}}{{\sin A}} - 2\cos (A + B)
Let us use the formula sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B to expand sin⁡(2A+B)\sin (2A + B).
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡Bsin⁡A−2cos⁡(A+B)\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B}}{{\sin A}} - 2\cos (A + B)
Now let us expand cos⁡(A+B)\cos (A + B) using the formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B.
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡Bsin⁡A−2(cos⁡Acos⁡B−sin⁡Asin⁡B)\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B}}{{\sin A}} - 2(\cos A\cos B - \sin A\sin B)
Now let us take LCM.
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡B−2sin⁡A(cos⁡Acos⁡B−sin⁡Asin⁡B)sin⁡A\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B - 2\sin A(\cos A\cos B - \sin A\sin B)}}{{\sin A}}
Let us simplify this.
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡B−2sin⁡Acos⁡Acos⁡B+2sin⁡Asin⁡Asin⁡B)sin⁡A\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B - 2\sin A\cos A\cos B + 2\sin A\sin A\sin B)}}{{\sin A}}
Now let us rewrite the above expression.
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡B−(2sin⁡Acos⁡A)cos⁡B+2sin⁡2Asin⁡B)sin⁡A\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B - (2\sin A\cos A)\cos B + 2{{\sin }^2}A\sin B)}}{{\sin A}}
Let us simplify the term 2sin⁡Acos⁡A2\sin A\cos Ausing the formulasin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A.
⇒sin⁡2Acos⁡B+cos⁡2Asin⁡B−sin⁡2Acos⁡B+2sin⁡2Asin⁡Bsin⁡A\Rightarrow \dfrac{{\sin 2A\cos B + \cos 2A\sin B - \sin 2A\cos B + 2{{\sin }^2}A\sin B}}{{\sin A}}
Now we can see that the first and the third term in the numerator are the same with a different sign so, they will get canceled.
⇒(cos⁡2A+2sin⁡2A)sin⁡Bsin⁡A\Rightarrow \dfrac{{(\cos 2A + 2{{\sin }^2}A)\sin B}}{{\sin A}}
Let us now expand cos⁡2A\cos 2Ausing the formulacos⁡2A=1−2sin⁡2A\cos 2A = 1 - 2{\sin ^2}A.
⇒(1−2sin⁡2A+2sin⁡2A)sin⁡Bsin⁡A\Rightarrow \dfrac{{(1 - 2{{\sin }^2}A + 2{{\sin }^2}A)\sin B}}{{\sin A}}
Now the term 2sin⁡2A2{\sin ^2}Awith different signs gets canceled.
⇒(1)sin⁡Bsin⁡A\Rightarrow \dfrac{{(1)\sin B}}{{\sin A}}
On simplifying this we get
⇒sin⁡Bsin⁡A\Rightarrow \dfrac{{\sin B}}{{\sin A}}
⇒LHS\Rightarrow LHS
⇒LHS=RHS\Rightarrow LHS = RHS

Note: In proving the statement we need to be choosier to select the identities so that we will get the function that is on the other side. Since we have more identities, we may end up expanding the trigonometry function in the wrong way, which means that we cannot make them like the function on the other side.