complex2 q (1 Viewer)

=)(= · Oct 13, 2022

part b i cant figure how to get the sum

Lith_30 · Oct 13, 2022

use the sum to product of pairs of roots to show that $\cos^2\left(\frac{\pi}{12}\right)+\cos^2\left(\frac{5\pi}{12}\right)=1$ then use the fact that $2\cos\left(\frac{\pi}{12}\right)\cos^2\left(\frac{5\pi}{12}\right)=\frac{1}{2}$ to find $\left(\cos\left(\frac{\pi}{12}\right)+\cos^2\left(\frac{5\pi}{12}\right)\right)^2=\frac{3}{2}$

and the rest should be easy

=)(= · Oct 13, 2022

would there be 15 pairs of prod to sum?

Lith_30 · Oct 13, 2022

=)(= said:
would there be 15 pairs of prod to sum?

yeah the roots would be $\pm\cos\left(\frac{\pi}{12}\right),\pm\cos\left(\frac{5\pi}{12}\right),\pm\frac{1}{\sqrt{2}}$

lets say you start off with $\cos\left(\frac{\pi}{12}\right)$ multiplying with $\pm\cos\left(\frac{5\pi}{12}\right),\pm\frac{1}{\sqrt{2}}$

and then $-\cos\left(\frac{\pi}{12}\right)$ multiplying with $\pm\cos\left(\frac{5\pi}{12}\right),\pm\frac{1}{\sqrt{2}}$

the product of those pairs of roots will cancel each other out when adding them to each other.

so basically you get all but three terms cancelling out, which is where the root multiplies by their negative equivalent ie $-\cos^2\left(\frac{\pi}{12}\right), -\cos^2\left(\frac{5\pi}{12}\right)\text{and}-\frac{1}{2}$

alternatively i think you might be able to do the question by letting $x=\cos^2(\theta)$

=)(= · Oct 13, 2022

ohh okay i get it thanks guys

Make a Difference – Donate to Bored of Studies!

complex2 q (1 Viewer)

=)(=

Active Member

Lith_30

o_o

=)(=

Active Member

Lith_30

o_o

=)(=

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)