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

Answer

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Hint – In this questions we have been given two equations and we need to find the value of$\left| {\sin \left( {A - B} \right)} \right|$. 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

$\sqrt 2 \cos A = \cos B + {\cos ^3}B$…………………….. (1)

$\sqrt 2 \sin A = \sin B - {\sin ^3}B$……………………… (2)

Now we have to find out the value of $\left| {\sin \left( {A - B} \right)} \right|$

So, divide equation (2) from equation (1) we have,

\[\dfrac{{\sqrt 2 \sin A}}{{\sqrt 2 \cos A}} = \dfrac{{\sin B - {{\sin }^3}B}}{{\cos B + {{\cos }^3}B}}\]

Now simplify the above equation we have,

\[\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B\left( {1 - {{\sin }^2}B} \right)}}{{\cos B\left( {1 + {{\cos }^2}B} \right)}}\]

Now as we know $\left( {1 - {{\sin }^2}\theta = {{\cos }^2}\theta } \right)$ so, use this property in above equation we have,

\[\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B{{\cos }^2}B}}{{\cos B \times \left( {1 + {{\cos }^2}B} \right)}} = \dfrac{{\sin B\cos B}}{{1 + {{\cos }^2}B}}\]

$ \Rightarrow \sin A\left( {1 + {{\cos }^2}B} \right) = \cos A\cos B\sin B$

$ \Rightarrow \sin A = \cos A\cos B\sin B - \sin A{\cos ^2}B$

$ \Rightarrow \sin A = \cos B\left( {\cos A\sin B - \sin A\cos B} \right)$

Now as we know $\cos A\sin B - \sin A\cos B = \sin \left( {B - A} \right) = - \sin \left( {A - B} \right)$ so, use this property in above equation we have,

$ \Rightarrow \sin A = \cos B\left( { - \sin \left( {A - B} \right)} \right)$

$ \Rightarrow \sin \left( {A - B} \right) = - \dfrac{{\sin A}}{{\cos B}}$

Now apply the modulus in above equation we have,

$ \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

$ \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.

“Complete step-by-step answer:”

Given equation is

$\sqrt 2 \cos A = \cos B + {\cos ^3}B$…………………….. (1)

$\sqrt 2 \sin A = \sin B - {\sin ^3}B$……………………… (2)

Now we have to find out the value of $\left| {\sin \left( {A - B} \right)} \right|$

So, divide equation (2) from equation (1) we have,

\[\dfrac{{\sqrt 2 \sin A}}{{\sqrt 2 \cos A}} = \dfrac{{\sin B - {{\sin }^3}B}}{{\cos B + {{\cos }^3}B}}\]

Now simplify the above equation we have,

\[\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B\left( {1 - {{\sin }^2}B} \right)}}{{\cos B\left( {1 + {{\cos }^2}B} \right)}}\]

Now as we know $\left( {1 - {{\sin }^2}\theta = {{\cos }^2}\theta } \right)$ so, use this property in above equation we have,

\[\dfrac{{\sin A}}{{\cos A}} = \dfrac{{\sin B{{\cos }^2}B}}{{\cos B \times \left( {1 + {{\cos }^2}B} \right)}} = \dfrac{{\sin B\cos B}}{{1 + {{\cos }^2}B}}\]

$ \Rightarrow \sin A\left( {1 + {{\cos }^2}B} \right) = \cos A\cos B\sin B$

$ \Rightarrow \sin A = \cos A\cos B\sin B - \sin A{\cos ^2}B$

$ \Rightarrow \sin A = \cos B\left( {\cos A\sin B - \sin A\cos B} \right)$

Now as we know $\cos A\sin B - \sin A\cos B = \sin \left( {B - A} \right) = - \sin \left( {A - B} \right)$ so, use this property in above equation we have,

$ \Rightarrow \sin A = \cos B\left( { - \sin \left( {A - B} \right)} \right)$

$ \Rightarrow \sin \left( {A - B} \right) = - \dfrac{{\sin A}}{{\cos B}}$

Now apply the modulus in above equation we have,

$ \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

$ \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.

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