Question

(cos9x-cos15x)/(cos6x*sin3x)

Answer 1

$\frac{ - 2 \sin(12x) \sin( - 3x) }{ \cos(6x) \sin(3x) }$
$\frac{4 \sin(6x) \cos(6x) \sin(3x) }{ \cos(6x) \sin(3x) } = 4 \sin(6x)$

Answer 2

To simplify the expression $\frac{\cos(9x) - \cos(15x)}{\cos(6x) \cdot \sin(3x)}$, we'll use trigonometric identities. Here's the step-by-step solution:

Step 1: Apply the trigonometric identity $\cos(A) - \cos(B) = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$ to the numerator.

$\frac{\cos(9x) - \cos(15x)}{\cos(6x) \cdot \sin(3x)} = \frac{-2 \sin\left(\frac{9x+15x}{2}\right) \sin\left(\frac{9x-15x}{2}\right)}{\cos(6x) \cdot \sin(3x)}$

Step 2: Simplify the arguments of the sines.

$\frac{-2 \sin\left(\frac{24x}{2}\right) \sin\left(\frac{-6x}{2}\right)}{\cos(6x) \cdot \sin(3x)}$

$\frac{-2 \sin(12x) \sin(-3x)}{\cos(6x) \cdot \sin(3x)}$

Step 3: Use the identity $\sin(-A) = -\sin(A)$.

$\frac{2 \sin(12x) \sin(3x)}{\cos(6x) \cdot \sin(3x)}$

Step 4: Cancel out the common factor of $\sin(3x)$ in the numerator and denominator.

$\frac{2 \sin(12x)}{\cos(6x)}$

Step 5: Apply the trigonometric identity $\sin(2A) = 2\sin(A)\cos(A)$ to the numerator.

$\frac{2 \cdot 2\sin(6x)\cos(6x)}{\cos(6x)}$

Step 6: Cancel out the common factor of $\cos(6x)$ in the numerator and denominator.

$2 \cdot 2\sin(6x) = 4\sin(6x)$

So, the simplified expression is $4\sin(6x)$.