The problem consists of two parts. 1. Prove that $\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2$. $x$ is a real number, not an integer multiple of $\frac{\pi}{2}$.

1. Problem Description

The problem consists of two parts.

1. Prove that $\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2$. $x$ is a real number, not an integer multiple of $\frac{\pi}{2}$.

2. Express the following expressions in terms of $cos2x$:

a) $\frac{sin3x}{sinx} + \frac{cos3x}{cosx}$

b) $\frac{sin5x}{sinx} - \frac{cos5x}{cosx}$

2. Solution Steps

1. Prove $\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2$

We can combine the terms on the left side by finding a common denominator:

$\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = \frac{sin3xcosx - cos3xsinx}{sinxcosx}$

Using the trigonometric identity $sin(A-B) = sinAcosB - cosAsinB$, we have:

$\frac{sin3xcosx - cos3xsinx}{sinxcosx} = \frac{sin(3x-x)}{sinxcosx} = \frac{sin2x}{sinxcosx}$

Using the trigonometric identity $sin2x = 2sinxcosx$, we have:

$\frac{sin2x}{sinxcosx} = \frac{2sinxcosx}{sinxcosx} = 2$

Therefore, $\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2$.

2. Express in terms of $cos2x$

a) $\frac{sin3x}{sinx} + \frac{cos3x}{cosx}$

$\frac{sin3x}{sinx} + \frac{cos3x}{cosx} = \frac{sin3xcosx + cos3xsinx}{sinxcosx}$

Using the trigonometric identity $sin(A+B) = sinAcosB + cosAsinB$, we have:

$\frac{sin3xcosx + cos3xsinx}{sinxcosx} = \frac{sin(3x+x)}{sinxcosx} = \frac{sin4x}{sinxcosx}$

Since $sin4x = 2sin2xcos2x = 4sinxcosxcos2x$, we have:

$\frac{sin4x}{sinxcosx} = \frac{4sinxcosxcos2x}{sinxcosx} = 4cos2x$

b) $\frac{sin5x}{sinx} - \frac{cos5x}{cosx}$

$\frac{sin5x}{sinx} - \frac{cos5x}{cosx} = \frac{sin5xcosx - cos5xsinx}{sinxcosx}$

Using the trigonometric identity $sin(A-B) = sinAcosB - cosAsinB$, we have:

$\frac{sin5xcosx - cos5xsinx}{sinxcosx} = \frac{sin(5x-x)}{sinxcosx} = \frac{sin4x}{sinxcosx}$

Since $sin4x = 2sin2xcos2x = 4sinxcosxcos2x$, we have:

$\frac{sin4x}{sinxcosx} = \frac{4sinxcosxcos2x}{sinxcosx} = 4cos2x$

3. Final Answer

1. $\frac{sin3x}{sinx} - \frac{cos3x}{cosx} = 2$

2. a) $\frac{sin3x}{sinx} + \frac{cos3x}{cosx} = 4cos2x$

b) $\frac{sin5x}{sinx} - \frac{cos5x}{cosx} = 4cos2x$