# Thread: Derivative of this trig function

1. ## Derivative of this trig function

Prove that

$\frac{d^n}{dx^n}(\sin^4x+\cos^4x)=4^{n-1}\cos(4x+\frac{n\pi}{2})$

The first thing to do, obviously, is to find the first derivative, which is

$4\sin^3x\cos x-4\cos^3x\sin x$

First I tried factoring, then using a couple of trigonometric identities.

$
-4\sin x\cos x(\cos^2x-\sin^2x)=-4\sin x\cos x(\cos(2x))=-4\sin x\cos x(2\cos^2x-1)
$

None of this, however, gets me any closer to $4^{n-1}\cos(4x+\frac{n\pi}{2})$. I differentiated a second time and got

-4(\cos^4x+\sin^4x-24\sin^2x\cos^2x)

I haven't double-checked my math on this last step but I'm pretty sure I'm right, but, regardless, it's not any closer either! So, I ask you, am i doing something wrong or just trying to follow the wrong steps?

2. $\displaystyle \sin^4{x}+\cos^4{x} = \frac{1}{4}\left(\cos{4x}+3\right)$, so $f'(x) = -\sin{4x} = \cos\left(4x+\frac{\pi}{2}\right)$.

3. u can prove that by induction let

$\displaystyle P_k : \frac{d^k}{dx^k} (\sin ^4 x + \cos ^4 x ) = 4^{n-1} \cos \right(4x + \frac{k\pi}{2} \left)$

for k= 1

$-4\sin x \cos x ( \cos 2x ) = -2\sin 2x (\cos 2x ) = - \sin 4x$
left hand side

$4^0 \cos (4x + \frac{\pi}{2} ) = \cos 4x \cos \frac{\pi}{2} - \sin 4x \sin \frac{pi}{2} = - \sin 4x$
same as the left hand side

now suppose P is true for n
want P is true for k = n+1 try it
$\displaystyle P_n : \frac{d^n}{dx^n} (\sin ^4 x + \cos ^4 x ) = 4^{n-1} \cos \right(4x + \frac{n\pi}{2} \left)$

can u continue ?

4. Originally Posted by TheCoffeeMachine
$\displaystyle \sin^4{x}+\cos^4{x} = \frac{1}{4}\left(\cos{4x}+3\right)$, so $f'(x) = -\sin{4x} = \cos\left(4x+\frac{\pi}{2}\right)$.
I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?

5. Originally Posted by Chokfull
I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?
Trig identities see here

List of trigonometric identities - Wikipedia, the free encyclopedia

6. Indeed...
Originally Posted by Chokfull
I don't get that first step...how do you get 1/4 (cos 4x+3) from sin^4 x+cos^4 x?
$\sin^4{x}+\cos^4{x} = \left(\sin^2{x}+\cos^2{x}\right)^2-2\left(\sin{x}\cos{x}\right)^2 = 1-2\left(\frac{1}{2}\sin{2x}\right)^2$

$\displaystyle 1-\frac{1}{2}\sin^2{2x} = \frac{2-\sin^2{2x}}{2} = \frac{2-\left(\frac{1-\cos{4x}}{2}\right)}{2} = \frac{\cos{4x}+4-1}{4} = \frac{1}{4}\left(\cos{4x}+3\right).$

Or if you are comfortable/allowed to use tables at leisure, just add the two reductions from the list:

$\displaystyle \sin^4{x}+\cos^4{x} = \left(\frac{3 - 4 \cos{2x} + \cos{4x}}{8}\right)+\left(\frac{3 + 4 \cos{2x} + \cos{4x}}{8}\right) = \frac{1}{4}\left(\cos{4x}+3\right)$.

7. ## Derivative of this trig function

Originally Posted by Chokfull
Prove that

$\frac{d^n}{dx^n}(\sin^4x+\cos^4x)=4^{n-1}\cos(4x+\frac{n\pi}{2})$

The first thing to do, obviously, is to find the first derivative, which is

$4\sin^3x\cos x-4\cos^3x\sin x$

First I tried factoring, then using a couple of trigonometric identities.

$
-4\sin x\cos x(\cos^2x-\sin^2x)=-4\sin x\cos x(\cos(2x))=\ \dots
$

Continuing from where I cut off your work ...

$= -2(\sin(2 x))(\cos(2x))$

$= -\sin(4 x)$

Shifting $\cos x$ by ${\pi\over2}$ to the left gives $-\sin x$, so $= -\sin(4 x)=\cos(4x+{\pi\over2})$.

So it works for $n=1$.

Now, what if you take the derivative of $\cos(4x+{\pi\over2})$ with respect to x?

8. thanks everyone