Simple differentiation problem, brainfart

[James889]

Hai,

I have the easiest problem but I am stuck at the last step, simplification.

\frac{cos3x+sin3x}{cos3x-sin3x}

\begin{aligned}<br /> F(x) = cos(3x)+sin(3x)\\<br /> F&#039;(x) = -3sin(3x)+3cos(3x) \\<br /> G(x) = cos(3x)-sin(3x) \\<br /> G&#039;(x) = -3sin(3x)-3cos(3x)\end{aligned}<br />

This gives \frac{(-3sin(3x)+3cos(3x))*(cos(3x)-sin(3x))+(cos(3x)+sin(3x))*(cos(3x)-sin(3x))}{(cos(3x)-sin(3x))^2}

But how would i simplify this?

 

[djeitnstine]

Expand the quadratics and look for identities. Already I can find sin^2 + cos^2 identities on both the top and bottom.

 

[AUMathTutor]

Also, it will help to remember that sin(u)cos(u) = (1/2)sin(2u)

 

[James889]

Okay, after some more trying:

I was able to reduce the first part(-3sin(3x)+3cos(3x))(cos(3x)-sin(3x))
Down to:
-3sin(3x)*cos(3x) + 3 -3cos(3x)*sin(3x)

And the second part:(cos(3x)+sin(3x))*(cos(3x)-sin(3x))
Down to:
cos(3x)^{2}-sin(3x)^{2}-2sin(3x)cos(3x)

Is that any better?

 

[djeitnstine]

Okay, after some more trying:

I was able to reduce the first part(-3sin(3x)+3cos(3x))(cos(3x)-sin(3x))
Down to:
-3sin(3x)*cos(3x) + 3 -3cos(3x)*sin(3x)

And the second part:(cos(3x)+sin(3x))*(cos(3x)-sin(3x))
Down to:
cos(3x)^{2}-sin(3x)^{2}-2sin(3x)cos(3x)

Is that any better?

Take note of AUMathtutor's comment

 

[AUMathTutor]

Also, naturally,

cos(u)^{2} - sin(u)^{2} = cos(2u)