Differentiation Rules of Sinusoidal Functions

chudzoik
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Homework Statement


f(x) = cos2x - sin2x

The Attempt at a Solution


f'(x) = (2cosx)(-sinx) - (2sinx)(cosx)
f'(x) = -2cosxsinx - 2sinxcosx

This is what I think the answer should be, but the back of the book says otherwise. I need help identifying what I did wrong.
 
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chudzoik said:

Homework Statement


f(x) = cos2x - sin2x

The Attempt at a Solution


f'(x) = (2cosx)(-sinx) - (2sinx)(cosx)
f'(x) = -2cosxsinx - 2sinxcosx

This is what I think the answer should be, but the back of the book says otherwise. I need help identifying what I did wrong.
... And what would that otherwise be?

Perhaps something involving sine and/or cosine of 2x ?

... or just a simplified version of what you have?

-2ab - 2ba = -4ab ?
 
SammyS said:
... And what would that otherwise be?

Perhaps something involving sine and/or cosine of 2x ?

... or just a simplified version of what you have?

-2ab - 2ba = -4ab ?

It says the answer should be f'(x) = -2sin2x.
 
chudzoik said:
It says the answer should be f'(x) = -2sin2x.

Recall this identity: 2sin(x)cos(x) = sin(2x)
 
Oh wow forgot entirely about that. These kinds of mistakes will be the end of me. :redface: Thanks for the help.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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