341right. your solution please?
10 Answers
341Its a circle. I know you will curse me for not providing the solution. But, you could try to prove this and also I want the students to have a go at it.
1thanks sir , i wanted to verify as answer is not given
sir is it a circle of radius 3 and centre shifted by two units in + x axis
?
1wont jus multiplying and dividing with conjugate of teh complex no 2+cosθ+isinθ help....
after that put z=x+iy and then equate real and imaginary parts.....and then eliminate cosθ from both eq to get an eq in x and y
341z = \frac{3}{2+\cos \theta + i \sin \theta} \Rightarrow \frac{3}{z} -2 = \cos \theta + i \sin \theta
\Rightarrow \left|\frac{3}{z} -2 \right|= 1 \Rightarrow |z| = 2 \left|z - \frac{3}{2} \right|
Thus, we are looking for points whose distances from (0,0) and (3/2,0) are in a fixed ratio which is nothing but a circle
(That's all they need, radius centre etc. are extraneous info)
1thanks sir :)
sir once i had seen u solving this question
p=({1+cos\frac{\pi}{10}})({1+cos\frac{3\pi}{10}})({1+cos\frac{7\pi}{10}})({1+cos\frac{9\pi}{10}})
using complex numbers + viete's formula
but i have lost that thread
sir , can u solve it once again plz
21for 2nd q u got to prove this
cos(n\theta) =2^{n-1}\left(cos\theta -cos\frac{\pi }{2n} \right)\left(cos\theta -cos\frac{3\pi }{2n} \right)..........\left(cos\theta -cos\frac{(2n-1)\pi }{2n} \right)
for this q
put θ=pie
n=5
ans 124