Rewrite with only sinx & cosx

Question

sin2x-cos3x

Answer 1

Consider the following identities:

 

sin(2x) = 2sinxcosx (double angle)

 

and

 

cos(2x) = 2cos²x-1 (double angle)

 

and

 

cos(a+b) = cos(a)cos(b) - sin(a)sin(b) (sum/difference)

 

and sin²x = 1-cos²x (Pythagorean)

 

NOW:

First, replace "3x" with "3x+x"

1) sin(2x) - cos(2x + x)

 

Utilize the sum formula for cos

2) sin (2x) - cos(2x)cos(x) - sin(2x)sin(x)

 

Utilize the double angle formulas for both of the sin(2x) and the cos(2x)

3) 2sinxcosx - (2cos²x-1)(cosx) - (2sinxcosx)(sinx)

 

Simplify a bit by removing the cosx

4) cosx [ 2sinx - (2cos²x-1) - 2sin²x ]

 

Distribute the minus sign in front of (2cos²x-1)

5) cosx [ 2sinx - 2cos²x + 1- 2sin²x]

 

Utilize the Pythagorean identity to replace sin²x

6) cosx [ 2sinx - 2cos²x + 1 - 2(1-cos²x) ]

 

Distribute the -2 in front of (1-cos²x)

7) cosx [ 2sinx - 2cos²x + 1 - 2 + 2cos²x]

 

Combine like terms (+1-2 = -1 and -2cos²x+2cos²x=0)

8) cosx [ 2sinx - 1]

 

 

There you have it.

Answer 2

Start with the formulas for double and triple angles:

 

sin 2x = 2 sin x cos x

cos 3x = 4cos³ x - 3cos x

 

So sin 2x - cos 3x =

2 sin x cosx - (4cos³ x - 3cos x) =

2 sin x cosx - 4 cos³ x + 3 cos x =

cos x (2 sin x - 4 cos² x + 3)

 

Is that what you were looking for?  There are other ways to write this, for example:

 

cos x (2 sin x - 4 cos² x + 3) =

cos x (2 sin x + 3 - 4 cos² x) =

cox x (2 sin x + 4 - 4 cos² x - 1) =

cox x (2 sin x + 4(1 - cos² x) - 1) =

cox x (2 sin x + 4 sin² x - 1)