# Thread: proof by induction ...

1. ## proof by induction ...

I am asked to prove the following by induction...

y=(e^x)sinx => d(^n)y/dx = [2^(n/2)] (e^x) [sin(x+npi/4)]

unfortunately I m trying to establish that this is true for n=1 and I am stuck already.

I get dy/dx = (e^x)(sinx-cosx) and not much passed that...

I have tried to square both sides and it gets me as far as...

dy/dx = [2(1/2)](e^x)[1/2 - sinxcosx]

....?

2. First of all we have

$\sin a+\cos a=\sin a+\cos\left(\frac{\pi}{2}-a\right)=2\sin\frac{\pi}{4}\cos\left(\frac{\pi}{4}-a\right)=$

$=\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sqrt{2}\sin\left(a+\frac{\ pi}{4}\right)$

Now, for n=1:

$(e^x\sin x)'=e^x(\sin x+\cos x)=\sqrt{2}e^x\sin\left(x+\frac{\pi}{4}\right)$

Suppose that $(e^x\sin x)^{(n)}=2^{\frac{n}{2}}e^x\sin\left(x+\frac{n\pi} {4}\right)$

We have to prove that $(e^x\sin x)^{(n+1)}=2^{\frac{n+1}{2}}e^x\sin\left(x+\frac{( n+1)\pi}{4}\right)$

$(e^x\sin x)^{(n+1)}=\left[(e^x\sin x)^{(n)}\right]'=\left[2^{\frac{n}{2}}e^x\sin\left(x+\frac{n\pi}{4}\right )\right]'=$

$=2^{\frac{n}{2}}e^x\left[\sin\left(x+\frac{n\pi}{4}\right)+\cos\left(x+\fra c{n\pi}{4}\right)\right]=$

$=2^{\frac{n}{2}}e^x\sqrt{2}\sin\left(x+\frac{n\pi} {4}+\frac{\pi}{4}\right)=2^{\frac{n+1}{2}}e^x\sin\ left(x+\frac{(n+1)\pi}{4}\right)$

3. really thanks! i will study it very carefully