Question

Let f(x)=3-2cos(x)sin(x). Give f ' (pi/4)

Answer 1

I should be using the product rule here, correct?

Answer 2

Yes

Answer 3

What I did was first try to find the derivative of x. I got f(x)=3-[2cos(x)*cos(x)+sin(x)-2sin(x)]

Answer 4

After working that out, I end up with 3-2cos^2(x)-2sin^2(x), which gives 3 when I plug in (pi/4)

Answer 5

But it is supposed to be 0, apparently.

Answer 6

\[f(x)=3-2\cos{x}\sin{x}=3-\sin{2x}\\f'(x)=-2\cos{2x}\]

Answer 7

\[f(x)=3-2\cos(x)\sin(x)\]
\[\frac{df(x)}{dx}=0-2[\frac{dsin(x)}{dx}\cos(x)+\frac{dcos(x)}{dx}\sin(x)]\]

Answer 8

\[f'(\frac\pi4)=-2\cos\frac\pi2=0\]

Answer 9

Oh, is that a trig identity I missed?

Answer 10

Yes. oldrin's method id best if you know that, otherwise- brute force works with differentiation.

Answer 11

Indeed!\[\sin{2x}=\sin{x+x}=\sin{x}\cos{x}+\sin{x}\cos{x}=2\sin{x}\cos{x}\]

Answer 12

We haven't learned Oldrin's method yet, I don't think. Thanks for the help, guys.

Answer 13

It comes from\[e^{\sqrt{-1}x}=\cos(x)+\sqrt{-1}\sin(x)\]

Answer 14

oldrin's identity, that is.

Answer 15

Because replacing x with 2x in the left term can be written as squaring the left term.

Answer 16

Ah, thanks. I might have to youtube that.

Answer 17

Euler's equation