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Question: If the value of \(\sqrt 2 \cos A = \cos B + {\cos ^3}B\) and \(\sqrt 2 \sin A = \sin B - {\sin ^3}B\...

If the value of 2cos⁡A=cos⁡B+cos⁡3B\sqrt 2 \cos A = \cos B + {\cos ^3}B and 2sin⁡A=sin⁡B−sin⁡3B\sqrt 2 \sin A = \sin B - {\sin ^3}B. Then find the value of ∣sin⁡(A−B)∣\left| {\sin \left( {A - B} \right)} \right|.

Explanation

Solution

Hint – In this questions we have been given two equations and we need to find the value of∣sin⁡(A−B)∣\left| {\sin \left( {A - B} \right)} \right|. So divide the two given questions and then using trigonometric identities solve the right hand side part. Since we have to tell the value of a quantity in modulus so after obtaining the value take modulus both the sides. This will help get the right answer.

“Complete step-by-step answer:”
Given equation is
2cos⁡A=cos⁡B+cos⁡3B\sqrt 2 \cos A = \cos B + {\cos ^3}B…………………….. (1)
2sin⁡A=sin⁡B−sin⁡3B\sqrt 2 \sin A = \sin B - {\sin ^3}B……………………… (2)
Now we have to find out the value of ∣sin⁡(A−B)∣\left| {\sin \left( {A - B} \right)} \right|
So, divide equation (2) from equation (1) we have,
2sin⁡A2cos⁡A=sin⁡B−sin⁡3Bcos⁡B+cos⁡3B\dfrac{{\sqrt 2 \sin A}}{{\sqrt 2 \cos A}} = \dfrac{{\sin B - {{\sin }^3}B}}{{\cos B + {{\cos }^3}B}}
Now simplify the above equation we have,
sin⁡Acos⁡A=sin⁡B(1−sin⁡2B)cos⁡B(1+cos⁡2B)\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B\left( {1 - {{\sin }^2}B} \right)}}{{\cos B\left( {1 + {{\cos }^2}B} \right)}}
Now as we know (1−sin⁡2θ=cos⁡2θ)\left( {1 - {{\sin }^2}\theta = {{\cos }^2}\theta } \right) so, use this property in above equation we have,
sin⁡Acos⁡A=sin⁡Bcos⁡2Bcos⁡B×(1+cos⁡2B)=sin⁡Bcos⁡B1+cos⁡2B\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B{{\cos }^2}B}}{{\cos B \times \left( {1 + {{\cos }^2}B} \right)}} = \dfrac{{\sin B\cos B}}{{1 + {{\cos }^2}B}}
⇒sin⁡A(1+cos⁡2B)=cos⁡Acos⁡Bsin⁡B\Rightarrow \sin A\left( {1 + {{\cos }^2}B} \right) = \cos A\cos B\sin B
⇒sin⁡A=cos⁡Acos⁡Bsin⁡B−sin⁡Acos⁡2B\Rightarrow \sin A = \cos A\cos B\sin B - \sin A{\cos ^2}B
⇒sin⁡A=cos⁡B(cos⁡Asin⁡B−sin⁡Acos⁡B)\Rightarrow \sin A = \cos B\left( {\cos A\sin B - \sin A\cos B} \right)
Now as we know cos⁡Asin⁡B−sin⁡Acos⁡B=sin⁡(B−A)=−sin⁡(A−B)\cos A\sin B - \sin A\cos B = \sin \left( {B - A} \right) = - \sin \left( {A - B} \right) so, use this property in above equation we have,
⇒sin⁡A=cos⁡B(−sin⁡(A−B))\Rightarrow \sin A = \cos B\left( { - \sin \left( {A - B} \right)} \right)
⇒sin⁡(A−B)=−sin⁡Acos⁡B\Rightarrow \sin \left( {A - B} \right) = - \dfrac{{\sin A}}{{\cos B}}
Now apply the modulus in above equation we have,
⇒∣sin⁡(A−B)∣=∣−sin⁡Acos⁡B∣\Rightarrow \left| {\sin \left( {A - B} \right)} \right| = \left| { - \dfrac{{\sin A}}{{\cos B}}} \right|
Now as we know modulus of any negative quantity is positive
⇒∣sin⁡(A−B)∣=sin⁡Acos⁡B\Rightarrow \left| {\sin \left( {A - B} \right)} \right| = \dfrac{{\sin A}}{{\cos B}}
So, this is the required answer.

Note – Whenever we face such types of problems the key concept is to have a good gist of the basic trigonometric identities some of them are being mentioned while performing the solution. Questions of such type are mostly formula based so having a good understanding of trigonometric identities helps getting on the right track to reach the solution.