0

Guest

Courses New

plz explain me this question … i will be very helpfull

plz explain me this question … i will be very helpfull
 

Question Image
Grade:12th pass

4 Answers

Amit
405 Points
6 years ago
Hi there.. In such ques in order to save time as well as your confidence you need to do by certain tricks.. you can do by actual method too but that would take time.. for this ques replace n=2 in ques as well as in answer .. just simplify a bit to get answer A.. for any doubts further you can add comment here.
sohan goswami
23 Points
6 years ago
how to add comment ,there is no such word as comment over here,so i am writing in the answer column , okay thank u for that
Amit
405 Points
6 years ago
I mean to say to write in answer column only. So sorry for using the word comment instead of that...Regards ..
Samyak Jain
333 Points
5 years ago
Let S = 1 + nC1 cos\theta + nC2 cos2\theta + ….................. + nCn cosn\theta
Now, (1 + x)n = 1 + nC1 x + nC2 x2 + ….................. + nCn xn
Put x = cos\theta + i sin\theta 
\therefore (1 + cos\theta + i sin\theta)n = 1 + nC1 (cos\theta + i sin\theta) + nC2 (cos\theta + i sin\theta)2 + …........ + nCn (cos\theta + i sin\theta)n    ...(1)
We know that 1 + cos\theta = 2cos2(\theta/2) , sin\theta = 2sin(\theta/2) cos(\theta/2)  & 
(cos\theta + i sin\theta)n = cos n\theta + i sin n\theta .
\therefore (1) is
[ 2cos2(\theta/2) + i 2sin(\theta/2) cos(\theta/2)]n = 1 + nC1 (cos\theta + i sin\theta) + nC2 (cos2\theta + i sin2\theta) + …........ + nCn (cosn\theta + i sinn\theta)
[ 2cos(\theta/2) {cos(\theta/2) + i sin(\theta/2)} ]n = 1 + nC1 cos\theta + nC1 i sin\theta + nC2 cos2\theta + nC2 i sin2\theta + …... + nCn cosn\theta + nCn i sinn\theta
i.e.  {2cos(\theta/2)}n {cos(n\theta/2) + i sin(n\theta/2)} = 1 + nC1 cos\theta + nC2 cos2\theta +  …... + nCn cosn\theta 
                                                                     + nC1 i sin\theta + nC2 i sin2\theta + ….... + nCn i sinn\theta
{2cos(\theta/2)}n cos(n\theta/2) + i{2cos(\theta/2)}n sin(n\theta/2) = 1 + nC1 cos\theta + nC2 cos2\theta +  …... + nCn cosn\theta 
                                                                               + i(nC1 sin\theta + nC2 sin2\theta + ….... + nCn sinn\theta)
Compare real and imaginary parts in the above equation.
\therefore 1 + nC1 cos\theta + nC2 cos2\theta + ….................. + nCn cosn\theta = {2cos(\theta/2)}n cos(n\theta/2)
&  nC1 sin\theta + nC2 sin2\theta + ….... + nCn sinn\theta = {2cos(\theta/2)}n sin(n\theta/2)

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More