Question

Prove $\sin 3x{\sin ^3}x + \cos 3x{\cos ^3}x = {\cos ^3}2x$.

Answer 1

Hint: Here we can solve the Left hand side of the given proof to get the required Right hand side. We can substitute the value of ${\sin ^3}x$ and ${\cos ^3}x$ from the formula $\sin 3x = 3\sin x - 4{\sin ^3}x$ and $\cos 3x = 4{\cos ^3}x - 3\cos x$.

Complete Step by Step Solution:
Here we are given to prove that $\sin 3x{\sin ^3}x + \cos 3x{\cos ^3}x = {\cos ^3}2x$
So here we can proceed by solving first the left hand side of the given equation which we need to prove.
For this we must know the general formula of $\sin 3x = 3\sin x - 4{\sin ^3}x$ and $\cos 3x = 4{\cos ^3}x - 3\cos x$.
From these formulas we can calculate the value of ${\sin ^3}x$ and ${\cos ^3}x$ which we can substitute in the given LHS and then we can simplify the LHS of the given proof to get the RHS of the proof.
Hence let us proceed by finding the value of ${\sin ^3}x$ and ${\cos ^3}x$
We have the formula which is:
$\sin 3x = 3\sin x - 4{\sin ^3}x$
From this formula we can get that:
${\sin ^3}x = \dfrac{1}{4}\left( {3\sin x - \sin 3x} \right)$$ - - - - (1)$
We also have the formula:
$\cos 3x = 4{\cos ^3}x - 3\cos x$
We will get:
${\cos ^3}x = \dfrac{1}{4}\left( {3\cos x + \cos 3x} \right)$$ - - - - \left( 2 \right)$
Hence we can substitute the above two values in the LHS we will get:
$
   = \sin 3x{\sin ^3}x + \cos 3x{\cos ^3}x \\
   = (\sin 3x)\dfrac{1}{4}\left( {3\sin x - \sin 3x} \right) + (\cos 3x)\dfrac{1}{4}\left( {3\cos x + \cos 3x} \right) \\
 $
We can simplify it and write as:
$
   = \dfrac{1}{4}(\sin 3x)\left( {3\sin x - \sin 3x} \right) + \dfrac{1}{4}(\cos 3x)\left( {3\cos x + \cos 3x} \right) \\
   = \dfrac{1}{4}\left( {3\left( {\cos 3x\cos x + \sin 3x\sin x} \right) + \left( {{{\cos }^2}3x - {{\sin }^2}3x} \right)} \right) - - \left( 3 \right) \\
 $
Now we know the formula that:
$\cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)$
Hence we can say that:
$\cos 3x\cos x + \sin 3x\sin x = \cos \left( {3x - x} \right) = \cos 2x - - - - - \left( 4 \right)$
Also we know the formula that:
${\cos ^2}A - {\sin ^2}A = \cos 2A$
Hence we can also say that:
${\cos ^2}3x - {\sin ^2}3x = \cos 2\left( {3x} \right) = \cos \left( {6x} \right) - - - - \left( 5 \right)$
Let us substitute the values that we have got in equation (4) and equation (5) in the equation (3), we will get:
LHS$ = \dfrac{1}{4}\left( {3\left( {\cos 3x\cos x + \sin 3x\sin x} \right) + \left( {{{\cos }^2}3x - {{\sin }^2}3x} \right)} \right)$
$ = \dfrac{1}{4}\left( {3\cos 2x + \cos 6x} \right)$
Now let us take $2x = A$
We will get:
$ = \dfrac{1}{4}\left( {3\cos 2x + \cos 6x} \right)$
$ = \dfrac{1}{4}\left( {3\cos A + \cos 3A} \right)$
We know the formula ${\cos ^3}A = \dfrac{1}{4}\left( {3\cos A + \cos 3A} \right)$
So we get that:
$\dfrac{1}{4}\left( {3\cos A + \cos 3A} \right) = {\cos ^3}A$
Now we can put the value of $A = 2x$ we will get:

$\dfrac{1}{4}\left( {3\cos 2x + \cos 6x} \right) = {\cos ^3}2x$${\text{ = RHS}}$

Note:
Here we must know the formula of all the trigonometric formulas and when to apply which formula. We must know the general formulae of all the trigonometric functions because without them the problem is difficult to solve.