Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

If $\alpha $ and $\beta$ are the roots of the equation $ x^{2}-2px+\left ( p^{2} + q^{2}\right )=0 $ and $ tan \; \theta =\large \frac{q}{y+p} $ show that $ \large\frac{\left (y+\alpha \right )^{n}-\left ( y+\beta \right )^{n}}{\alpha -\beta }$ = $ q^{n-1}\large\frac{sin \;n\theta }{sin\;^{n}\theta }$

n N

1 Answer

Comment
A)
Toolbox:
  • From De moivre's theorem we have
  • (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
  • (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
  • (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
  • (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
  • $e^{i\theta}=\cos\theta+i\sin\theta$
  • $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
$\alpha$ and $\beta$ are the roots of the equation $x^2-2px+(p^2+q^2)=0$
Solving $x=\large\frac{2p\pm\sqrt{4p^2-4(p^2+q^2)}}{2}$
$\Rightarrow \large\frac{2p\pm i2q}{2}$
$\Rightarrow p\pm iq$
Step 2:
Let $\alpha=p+iq,$ then $\beta=p-iq\Rightarrow \alpha-\beta=2qi$
We also have $\tan\theta=\large\frac{q}{y+p}$
$\Rightarrow y+p=\large\frac{q}{\tan\theta}=\large\frac{q\cos\theta}{\sin\theta}$
Step 3:
Now $\large\frac{(y+\alpha)^n-(y+\beta)^n}{\alpha-\beta}=\large\frac{(y+p+iq)^n-(y+p-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{(\Large\frac{q\cos\theta}{\sin\theta}\normalsize+iq)^n-(\Large\frac{q\cos\theta}{\sin\theta}\normalsize-iq)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{(\cos \theta+i\sin\theta)^n-(\cos\theta-i\sin\theta)^n}{2qi}$
$\qquad\qquad=\large\frac{q^n}{\sin^n\theta}\large\frac{\cos n\theta+i\sin n\theta-\cos n\theta+i\sin n\theta}{2qi}$
$\qquad\qquad=\large\frac{q^n}{2qi\sin^n\theta}$$2i\sin n\theta$
$\qquad\qquad=\large\frac{q^n}{q\sin^n\theta}$$\sin n\theta$
$\qquad\qquad=\large\frac{q^{n-1}}{\sin^n\theta}$$\sin n\theta$
Hence proved.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...