# How do you verify the identity (cos3beta)/cosbeta=1-4sin^2beta?

Jan 14, 2017

$\cos \left(3 \beta\right) = \cos \left(2 \beta + \beta\right)$.

Use the sum formula $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ to expand:

LHS:

$\frac{\cos 2 \beta \cos \beta - \sin 2 \beta \sin \beta}{\cos} \beta$

Expand using the identities $\cos 2 x = 2 {\cos}^{2} x - 1$ and $\sin 2 x = 2 \sin x \cos x$:

$\frac{\left(2 {\cos}^{2} \beta - 1\right) \cos \beta - \left(2 \sin \beta \cos \beta\right) \sin \beta}{\cos} \beta$

$\frac{2 {\cos}^{3} \beta - \cos \beta - 2 {\sin}^{2} \beta \cos \beta}{\cos} \beta$

Use ${\sin}^{2} x + {\cos}^{2} x = 1$:

$\frac{2 {\cos}^{3} \beta - \cos \beta - 2 \left(1 - {\cos}^{2} \beta\right) \cos \beta}{\cos} \beta$

$\frac{2 {\cos}^{3} \beta - \cos \beta - 2 \left(\cos \beta - {\cos}^{3} \beta\right)}{\cos} \beta$

$\frac{2 {\cos}^{3} \beta - \cos \beta - 2 \cos \beta + 2 {\cos}^{3} \beta}{\cos} \beta$

$\frac{4 {\cos}^{3} \beta - 3 \cos \beta}{\cos} \beta$

$\frac{\cos \beta \left(4 {\cos}^{2} \beta - 3\right)}{\cos} \beta$

$4 {\cos}^{2} \beta - 3$

Switch into sine now using ${\cos}^{2} x = 1 - {\sin}^{2} x$:

$4 \left(1 - {\sin}^{2} \beta\right) - 3$

$4 - 4 {\sin}^{2} \beta - 3$

$1 - 4 {\sin}^{2} \beta$

Since the $L H S$ equals the $R H S$, we are done here.

Hopefully this helps!

Jan 14, 2017

See the Proof in the Explanation Section.

#### Explanation:

We will need $\cos 2 \beta = 1 - 2 {\sin}^{2} \beta$

$\cos 3 \beta = \underline{\cos 3 \beta + \cos \beta} - \cos \beta \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

Since, $\cos C + \cos D = 2 \cos \left(\frac{C + D}{2}\right) \cos \left(\frac{C - D}{2}\right)$

$\therefore , \text{ from } \left(1\right) , \cos 3 \beta = 2 \cos 2 \beta \cos \beta - \cos \beta ,$

$= \left(\cos \beta\right) \left(2 \cos 2 \beta - 1\right) ,$

$= \left(\cos \beta\right) \left\{2 \left(1 - 2 {\sin}^{2} \beta\right) - 1\right\} , i . e . ,$

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

$\Rightarrow \frac{\cos 3 \beta}{\cos} \beta = 1 - 4 {\sin}^{2} \beta$

Enjoy Maths.!