# Using the identity cos(A+B) = cosAcosB - sinAsinB, how do you prove that 1/4cos(3A) = cos^3A - 3/4cosA?

Jun 12, 2016

I'm going to work from the right hand side. For $\cos 3 A$, we know that:

$\textcolor{g r e e n}{\cos 3 A = \cos \left(A + 2 A\right) = \cos A \textcolor{h i g h l i g h t}{\cos 2 A} - \sin A \textcolor{h i g h l i g h t}{\sin 2 A}} .$

Next, recall that:

• $\textcolor{h i g h l i g h t}{\sin 2 A} = \sin A \cos A + \cos A \sin A = \textcolor{h i g h l i g h t}{2 \sin A \cos A}$
• $\textcolor{h i g h l i g h t}{\cos 2 A} = \cos A \cos A - \sin A \sin A = \textcolor{h i g h l i g h t}{{\cos}^{2} A - {\sin}^{2} A}$

So, we now have:

$\cos 3 A = \cos A \left(\textcolor{h i g h l i g h t}{{\cos}^{2} A - {\sin}^{2} A}\right) - \sin A \left(\textcolor{h i g h l i g h t}{2 \sin A \cos A}\right)$

$= {\cos}^{3} A - {\sin}^{2} A \cos A - 2 {\sin}^{2} A \cos A$

$= {\cos}^{3} A - 3 \textcolor{red}{{\sin}^{2}} A \cos A$

What I'm shooting for at this point is the result you get on the left hand side when you multiply the right hand side by $4$: $\textcolor{p u r p \le}{4 {\cos}^{3} A - 3 \cos A}$. That requires all functions to be $\cos$ here.

So, we have to reference the identity that you definitely know: $\textcolor{red}{{\sin}^{2} A + {\cos}^{2} A = 1}$. This gives us:

$= {\cos}^{3} A - 3 \left(\textcolor{red}{1 - {\cos}^{2} A}\right) \cos A$

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

$= {\cos}^{3} A - 3 \cos A + 3 {\cos}^{3} A$

$= \textcolor{p u r p \le}{4 {\cos}^{3} A - 3 \cos A}$

Almost there! Finally, since this was equal to $\cos 3 A$, we have to divide by $4$ to get:

$\frac{1}{4} \cos 3 A = \frac{1}{4} \cdot \left[4 {\cos}^{3} A - 3 \cos A\right]$

$\implies \textcolor{b l u e}{\frac{1}{4} \cos 3 A = {\cos}^{3} A - \frac{3}{4} \cos A}$

Jun 13, 2016

Using only the given identity of $\cos \left(A + B\right)$ the other given identity can be proved as follows.

#### Explanation:

General Discussion

The identity to be assumed is

$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B \ldots \left(1\right)$

As this is valid for all real values of A and B , we can put convenient values to get different identities as follows.

Method - I( A Tricky one)

Putting $B = - B$ in equation (1)

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

=>cos(A-B)=cosAcosB+sinAsinB…(1.1)

Adding (1) and (1.1) we get

cos(A+B)+cos(A-B)=2cosAcosB…(1.2)

Now Putting $B = 2 A$ in (1.2)

$\cos \left(A + 2 A\right) + \cos \left(A - 2 A\right) = 2 \cos A \cos 2 A$

$\cos 3 A + \cos A = 2 \cos A \cos 2 A$

cos3A=cosA(2cos2A-1)…(1.3)

Putting $B = A$ in (1.2)

$\cos \left(A + A\right) + \cos \left(A - A\right) = 2 \cos A \cos A$

$\implies \cos 2 A + \cos 0 = 2 {\cos}^{2} A$

cos2A=2cos^2A-1…(1.4)

Finally combining (1.4) with (1.3) we get

$\cos 3 A = \cos A \left(2 \left(2 {\cos}^{2} A - 1\right) - 1\right)$

$\implies \cos 3 A = 4 {\cos}^{3} A - 3 \cos A$

Re-arrenging and deviding bothsides by 4
${\cos}^{3} A - \frac{3}{4} \cos A = \frac{1}{4} \cos 3 A$
Proved

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method-II (A long one)

Putting $B = A$ in equation(1)

$\cos \left(A + A\right) = \cos A \cos A - \sin A \sin A$

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

Putting$B = 90 + A$in equation(1)

$\cos \left(A + 90 + A\right) = \cos A \cos \left(90 + A\right) - \sin A \sin \left(90 + A\right)$

$- \sin 2 A = - \sin A \cos A - \sin A \cos A$

$\implies \sin 2 A = 2 \sin A \cos A \ldots \left(3\right)$

Putting $B = - A$ in equation(1)

$\cos \left(A - A\right) = \cos A \cos \left(- A\right) - \sin A \sin \left(- A\right)$

$\implies {\cos}^{2} A + {\sin}^{2} A = 1. . . \left(4\right)$

Now finally putting $B = 2 A$ in equation (1)

$\cos \left(A + 2 A\right) = \cos A \cos 2 A - \sin A \sin 2 A$

inserting the relation (3)

$\cos 3 A = \cos A \cos 2 A - 2 \sin A \sin A \cos A$

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

Inserting relation (2)

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

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

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

=cosA(4cos^2A-3(cos^2A+sin^2A)

$= \cos A \left(4 {\cos}^{2} A - 3 \cdot 1\right) \text{ } r e l a t i o n \left(4\right)$

$\cos 3 A = 4 {\cos}^{3} A - 3 \cos A$

Re-arrenging and deviding bothsides by 4

$\therefore {\cos}^{3} A - \frac{3}{4} \cos A = \frac{1}{4} \cos 3 A$

Proved

Jun 14, 2016

RHS is
$\frac{1}{4} \cos \left(3 A\right) = \frac{1}{4} \cos \left(2 A + A\right)$
Using the given expression
$\implies \frac{1}{4} \left(\cos \left(2 A\right) \cos \left(A\right) - \sin \left(2 A\right) \sin A\right)$

1. Substituting the known expression for $\sin \left(2 A\right) = 2 \sin A \cos A$
2. Using given expression again for
$\cos \left(2 A\right) = \cos \left(A + A\right) = {\cos}^{2} A - {\sin}^{2} A$
3. and writing all terms in $\text{cosine}$ using ${\sin}^{2} A + {\cos}^{2} A = 1$ we obtain
$\frac{1}{4} \left(\left(2 {\cos}^{2} A - 1\right) \cos A - 2 {\sin}^{2} A \cos A\right)$

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