Prove: sinx-2sin3x+sin5x=2sinx(cos4x-cos2x )

A = sinx - sin3x,  

B = -sin3x + sin5x  

First A:  

The average of x and 3x is 2x, and they (x and 3x, that is) are each a distance of x from this average. That's fancy talk for:  

x = 2x-x,  

3x = 2x+x.  

So, A = sin(2x-x) - sin(2x+x)  

Using angle sum formulas:  

A = (sin2x cosx - cos2x sinx) - (sin2x cosx + cos2x sinx)  

A = -2 cos2x sinx  

Similarly,  

B = -sin(4x-x) + sin(4x+x)  

= -(sin4x cosx - cos4x sinx) + (sin4x cosx + cos4x sinx)  

B = 2 cos4x sinx  

Now,  

sinx - 2sin3x + sin5x = A+B = -2 cos2x sinx + 2 cos4x sinx  

= 2 sinx (cos4x - cos2x).  

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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