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# Prove that$\sin 3A.{\sin ^3}A + \cos 3A.{\cos ^3}A = {\cos ^3}2A$

Answer Verified
Hint: We cannot solve the Question directly , we have to use some identities, which identity is to be used. We can get a hint of it from the question itself like cos3A whose identity is known to us, simplify the equation and see if you can get something common out and proceed.

Complete step-by-step answer:

We know that
$\cos 3A = 4{\cos ^3}A - 3\cos A \\ \sin 3A = 3\sin A - 4{\sin ^3}A \\$

Using these in L.H.S
$\therefore$ The L.H.S
= ${\sin ^3}A\left( {3\sin A - 4{{\sin }^3}A} \right) + {\cos ^3}A\left( {4{{\cos }^3}A - 3\cos A} \right)$
= $3{\sin ^4}A - 4{\sin ^6}A + 4{\cos ^6}A - 3{\cos ^4}A$

Taking common
$= - 3\left( {{{\cos }^4}A - {{\sin }^4}A} \right) + 4\left( {{{\cos }^6}A - {{\sin }^6}A} \right) \\ = 4\left( {{{\left( {{{\cos }^2}A} \right)}^3} - {{\left( {{{\sin }^2}A} \right)}^3}} \right) - 3\left( {{{\left( {{{\cos }^2}A} \right)}^2} - {{\left( {{{\sin }^2}A} \right)}^2}} \right) \\$

Using basic algebraic formula,
$= 4\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {{{({{\cos }^2}A)}^2} + {{\left( {{{\sin }^2}A} \right)}^2} + {{\cos }^2}A{{\sin }^2}A} \right) - 3\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {{{\cos }^2}A + {{\sin }^2}A} \right)$

We know that ${\sin ^2}A + {\cos ^2}A = 1$,${\cos ^2}A - {\sin ^2}A = \cos 2A$
And taking common $\left( {{{\cos }^2}A - {{\sin }^2}A} \right)$
$= \cos 2A\left( {4{{\cos }^4}A + 4{{\sin }^4}A + 4{{\cos }^2}A{{\sin }^2}A - 3.1} \right)$

We can replace 1 by ${\left( {{{\sin }^2}A + {{\cos }^2}A} \right)^2}$
$= \cos 2A\left( {4{{\cos }^4}A + 4{{\sin }^4}A + 4{{\cos }^2}A{{\sin }^2}A - 3.{{\left( {{{\sin }^2}A + {{\cos }^2}A} \right)}^2}} \right)$$= \cos 2A\left( {4{{\cos }^4}A + 4{{\sin }^4}A + 4{{\cos }^2}A{{\sin }^2}A - 3.{{\left( {{{\sin }^4}A + {{\cos }^4}A + 2{{\sin }^2}A{{\cos }^2}A} \right)}^{}}} \right)$$= \cos 2A\left( {4{{\cos }^4}A + 4{{\sin }^4}A + 4{{\cos }^2}A{{\sin }^2}A - 3{{\sin }^4}A - 3{{\cos }^4}A - 6{{\cos }^2}A{{\sin }^2}A} \right)$
$= \cos 2A\left( {{{\cos }^4}A + {{\sin }^4}A - 2{{\cos }^2}A{{\sin }^2}A} \right) \\ = \cos 2A{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)^2} \\ = \cos 2A.{(\cos 2A)^2} \\ = {\cos ^3}2A \\$
=R.H.S

Note: Before solving such questions one needs to have a fine knowledge of trigonometry. About the properties of sine and cosines. Observe the question carefully the given terms will be like those in the properties, then only you will get to know what to apply.