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Question: If \( m{\sin ^2}A = {\cos ^2}\left( {A - B} \right) + {\cos ^2}B - 2\cos \left( {A - B} \right)\cos ...

If msin⁡2A=cos⁡2(A−B)+cos⁡2B−2cos⁡(A−B)cos⁡Acos⁡Bm{\sin ^2}A = {\cos ^2}\left( {A - B} \right) + {\cos ^2}B - 2\cos \left( {A - B} \right)\cos A\cos B . Find m.

Explanation

Solution

In this question, we will use the trigonometric identity cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B and square identity (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab to simplify the expression given LHS. We will also use the formula (cos⁡2A+sin⁡2A)=1\left( {{{\cos }^2}A + {{\sin }^2}A} \right) = 1 to get the final simplified expression. After this, we compare the RHS and LHS to get the value of m.

Complete step-by-step answer:
Given:- msin⁡2A=cos⁡2(A−B)+cos⁡2B−2cos⁡(A−B)cos⁡Acos⁡Bm{\sin ^2}A = {\cos ^2}\left( {A - B} \right) + {\cos ^2}B - 2\cos \left( {A - B} \right)\cos A\cos B
We know that
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B and also :
(a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab ----------- (1)
Applying this formula on RHS we get,
(cos⁡Acos⁡B+sin⁡Asin⁡B)2+cos⁡2B−2(cos⁡Acos⁡B+sin⁡Asin⁡B)cos⁡Acos⁡B{\left( {\cos A\cos B + \sin A\sin B} \right)^2} + {\cos ^2}B - 2\left( {\cos A\cos B + \sin A\sin B} \right)\cos A\cos B
On expanding the terms using identity given by equation 1, we get:
=cos⁡2Acos⁡2B+sin⁡2Asin⁡2B+2cos⁡Acos⁡Bsin⁡Asin⁡B+cos⁡2B−2cos⁡2Acos⁡2B−2sin⁡Asin⁡Bcos⁡Acos⁡B= {\cos ^2}A{\cos ^2}B + {\sin ^2}A{\sin ^2}B + 2\cos A\cos B\sin A\sin B + {\cos ^2}B - 2{\cos ^2}A{\cos ^2}B - 2\sin A\sin B\cos A\cos B
On cancelling the two equal terms with opposite sign, we have:
=cos⁡2Acos⁡2B+sin⁡2Asin⁡2B−2cos⁡2Acos⁡2B+cos⁡2B= {\cos ^2}A{\cos ^2}B + {\sin ^2}A{\sin ^2}B - 2{\cos ^2}A{\cos ^2}B + {\cos ^2}B
=sin⁡2Asin⁡2B−cos⁡2Acos⁡2B+cos⁡2B= {\sin ^2}A{\sin ^2}B - {\cos ^2}A{\cos ^2}B + {\cos ^2}B
On rearranging the terms, we have:
=cos⁡2B−cos⁡2Acos⁡2B+sin⁡2Asin⁡2B= {\cos ^2}B - {\cos ^2}A{\cos ^2}B + {\sin ^2}A{\sin ^2}B
On taking out the Cos⁡2B{\operatorname{Cos} ^2}B as common, we get:
=cos⁡2B(1−cos⁡2A)+sin⁡2Asin⁡2B= {\cos ^2}B\left( {1 - {{\cos }^2}A} \right) + {\sin ^2}A{\sin ^2}B

Now, we know that
(cos⁡2A+sin⁡2A)=1\left( {{{\cos }^2}A + {{\sin }^2}A} \right) = 1 -------(2)
⇒sin⁡2A=1−cos⁡2A\Rightarrow {\sin ^2}A = 1 - {\cos ^2}A ---------- (3)
Using equation 3, the above expression can be written as:
=cos⁡2Bsin⁡2A+sin⁡2Asin⁡2B= {\cos ^2}B{\sin ^2}A + {\sin ^2}A{\sin ^2}B
Now, on taking out the Sin⁡2A{\operatorname{Sin} ^2}A as common, we get:
=sin⁡2A(cos⁡2B+sin⁡2B)= {\sin ^2}A\left( {{{\cos }^2}B + {{\sin }^2}B} \right)
Using the identity given by equation 2, we have:
=sin⁡2A(1)= {\sin ^2}A\left( 1 \right)
=sin⁡2A= {\sin ^2}A
Since on comparing this simplified value of RHS with LHS, we get the value of m is 1
i.e. msin⁡3A=sin⁡3Am{\sin ^3}A = {\sin ^3}A
Thus m = 1

Note: In this type of question we just need to solve one side and then compare it to the other side. In case if the other side is also uncomplicated form then firstly simplify it then compare to get the value. Before solving these types of questions, you should remember the important trigonometric formulas and identities like Sin(A±B) = SinACosB±CosASinBSin\left( {A \pm B} \right){\text{ }} = {\text{ }}SinACosB \pm CosASinB .