Prove that cos^3Acos(3A)+sin^3A(sin3A)=cos^3 2A ?

Oct 25, 2017

Please refer to a Proof given in the Explanation Section.

Explanation:

Recall that,

cos3A=4cos^3A-3cosA, &, sin3A=3sinA-4sin^3A.

Sub.ing these, we have,

$\therefore \text{ The L.H.S.=} {\cos}^{3} A \left(4 {\cos}^{3} A - 3 \cos A\right) + {\sin}^{3} A \left(3 \sin A - 4 {\sin}^{3} A\right) ,$

$= 4 {\cos}^{6} A - 3 {\cos}^{4} A + 3 {\sin}^{4} A - 4 {\sin}^{6} A ,$

$= 4 \left({\cos}^{6} A - {\sin}^{6} A\right) - 3 \left({\cos}^{4} A - {\sin}^{4} A\right) ,$

$= 4 \left\{{\left({\cos}^{2} A\right)}^{3} - {\left({\sin}^{2} A\right)}^{3}\right\} - 3 \left\{{\left({\cos}^{2} A\right)}^{2} - {\left({\sin}^{2} A\right)}^{2}\right\} ,$

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

$- 3 \left({\cos}^{2} A - {\sin}^{2} A\right) \left({\cos}^{2} A + {\sin}^{2} A\right) ,$

$= \left({\cos}^{2} A - {\sin}^{2} A\right) \left[4 \left\{{\left({\cos}^{2} A\right)}^{2} + {\cos}^{2} A {\sin}^{2} A + {\left({\sin}^{2} A\right)}^{2}\right\} - 3 \cdot 1\right] ,$

$= \left({\cos}^{2} A - {\sin}^{2} A\right) \left[4 {\cos}^{4} A + 4 {\cos}^{2} A {\sin}^{2} A + 4 {\sin}^{4} A - 3 {\left({\cos}^{2} A + {\sin}^{2} A\right)}^{2}\right] ,$

$= \left({\cos}^{2} A - {\sin}^{2} A\right) \left[4 {\cos}^{4} A + 4 {\cos}^{2} A {\sin}^{2} A + 4 {\sin}^{4} A - 3 \left({\cos}^{4} A + 2 {\cos}^{2} A {\sin}^{2} A + {\sin}^{4} a\right)\right] ,$

$= \left({\cos}^{2} A - {\sin}^{2} A\right) \left[{\cos}^{4} A - 2 {\cos}^{2} A {\sin}^{2} A + {\sin}^{4} A\right] ,$

$= \left({\cos}^{2} A - {\sin}^{2} A\right) {\left({\cos}^{2} A - {\sin}^{2} A\right)}^{2} ,$

$= {\left({\cos}^{2} A - {\sin}^{2} A\right)}^{3} ,$

$= {\left(\cos 2 A\right)}^{3} ,$

$= {\cos}^{3} 2 A ,$

$\text{=The R.H.S.}$

Enjoy Maths.!

Oct 25, 2017

Kindly refer to a Proof in the Explanation.

Explanation:

Here is a Second Method to solve the Problem.

We know the following Identities :

$\cos 3 A = 4 {\cos}^{3} A - 3 \cos A , \mathmr{and} , \sin 3 A = 3 \sin A - 4 {\sin}^{3} A .$

$\therefore {\cos}^{3} A = \frac{1}{4} \left(\cos 3 A + 3 \cos A\right) \ldots \ldots . \left(1\right) , \mathmr{and} ,$

${\sin}^{3} A = \frac{1}{4} \left(3 \sin A - \sin 3 A\right) \ldots \ldots \ldots \ldots . \left(2\right) .$ Using these, we have,

$\text{The L.H.S=} \frac{1}{4} \left\{\left(\cos 3 A + 3 \cos A\right) \cos 3 A + \left(3 \sin A - \sin 3 A\right) \sin 3 A\right] ,$

$= \frac{1}{4} \left\{{\cos}^{2} 3 A + 3 \cos 3 A \cos A + 3 \sin A \sin 3 A - {\sin}^{2} 3 A\right\} ,$

$= \frac{1}{4} \left\{\left({\cos}^{2} 3 A - {\sin}^{2} 3 A\right) + 3 \left(\cos 3 A \cos A + \sin 3 A \sin A\right)\right\} ,$

$= \frac{1}{4} \left\{\cos \left(2 \times 3 A\right) + 3 \cos \left(3 A - A\right)\right\} ,$

$= \frac{1}{4} \left\{\cos \left(3 \times 2 A\right) + 3 \cos 2 A\right\} ,$

$= \frac{1}{4} \left(\cos 3 \theta + 3 \cos \theta\right) , w h e r e , \theta = 2 A ,$

$= {\cos}^{3} \theta \ldots \ldots \ldots . \left[\because , \left(1\right)\right] ,$

$= {\cos}^{3} 2 A ,$

$\text{=The R.H.S.}$

Enjoy Maths.!

Oct 25, 2017

$L H S = {\cos}^{3} A \cos 3 A + {\sin}^{3} A \sin 3 A$

$= \frac{1}{2} \left({\cos}^{2} A \cdot 2 \cos A \cos 3 A + {\sin}^{2} A \cdot 2 \sin A \sin 3 A\right)$

$= \frac{1}{2} \left[{\cos}^{2} A \left(\cos 4 A + \cos 2 A\right) + {\sin}^{2} A \left(\cos 2 A - \cos 4 A\right)\right]$

$= \frac{1}{2} \left[{\cos}^{2} A \cdot \cos 4 A + {\cos}^{2} A \cdot \cos 2 A + {\sin}^{2} A \cdot \cos 2 A - {\sin}^{2} A \cdot \cos 4 A\right]$

$= \frac{1}{2} \left[{\cos}^{2} A \cdot \cos 4 A - {\sin}^{2} A \cdot \cos 4 A + {\cos}^{2} A \cdot \cos 2 A + {\sin}^{2} A \cdot \cos 2 A\right]$

$= \frac{1}{2} \left[\left({\cos}^{2} A - {\sin}^{2} A\right) \cos 4 A + \left({\cos}^{2} A + {\sin}^{2} A\right) \cos 2 A\right]$

$= \frac{1}{2} \left[\cos 2 A \cdot \cos 4 A + \cos 2 A\right]$

$= \frac{1}{2} \left[\cos 2 A \left(\cos 4 A + 1\right)\right]$

$= \frac{1}{2} \cdot \cos 2 A \cdot 2 {\cos}^{2} 2 A$

$= {\cos}^{3} 2 A = R H S$