Algebra> plz explain me this question … i will be ...

plz explain me this question … i will be very helpfull

sohan goswami , 7 Years ago

Grade 12th pass

4 Answers

Amit

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.

Last Activity: 7 Years ago

sohan goswami

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

Last Activity: 7 Years ago

Amit

I mean to say to write in answer column only. So sorry for using the word comment instead of that...Regards ..

Last Activity: 7 Years ago

Samyak Jain

Let S = 1 + ⁿC₁cos $\theta$ + ⁿC₂cos2 $\theta$ + ….................. + ⁿC_n cosn $\theta$

Now, (1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂ x² + ….................. + ⁿC_n 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 + …........ + ⁿC_n(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 $\theta$ ) + …........ + ⁿC_n(cosn $\theta$ + i sinn $\theta$ )

[ 2cos( $\theta$ /2) {cos( $\theta$ /2) + i sin( $\theta$ /2)} ]ⁿ= 1 + ⁿC₁cos $\theta$ + ⁿC₁i sin $\theta$ + ⁿC₂cos2 $\theta$ + ⁿC₂ i sin2 $\theta$ + …... + ⁿC_ncosn $\theta$ + ⁿC_n i sinn $\theta$

i.e. {2cos( $\theta$ /2)}ⁿ{cos(n $\theta$ /2) + i sin(n $\theta$ /2)} = 1 + ⁿC₁cos $\theta$ + ⁿC₂cos2 $\theta$ + …... + ⁿC_ncosn $\theta$

+ ⁿC₁i sin $\theta$ + ⁿC₂ i sin2 $\theta$ + ….... + ⁿC_n i sinn $\theta$

{2cos( $\theta$ /2)}ⁿcos(n $\theta$ /2) + i{2cos( $\theta$ /2)}ⁿ sin(n $\theta$ /2) = 1 + ⁿC₁cos $\theta$ + ⁿC₂cos2 $\theta$ + …... + ⁿC_ncosn $\theta$

+ i(ⁿC₁sin $\theta$ + ⁿC₂ sin2 $\theta$ + ….... + ⁿC_n sinn $\theta$ )

Compare real and imaginary parts in the above equation.

$\therefore$ 1 + ⁿC₁cos $\theta$ + ⁿC₂cos2 $\theta$ + ….................. + ⁿC_n cosn $\theta$ = {2cos( $\theta$ /2)}ⁿcos(n $\theta$ /2)

& ⁿC₁sin $\theta$ + ⁿC₂ sin2 $\theta$ + ….... + ⁿC_n sinn $\theta$ = {2cos( $\theta$ /2)}ⁿ sin(n $\theta$ /2)

Last Activity: 7 Years ago