Complex help please (1 Viewer)

Aysce

Well-Known Member
Joined
Jun 24, 2011
Messages
2,394
Gender
Male
HSC
2012
Simplify leaving answers in exact values

I do not understand the solution in Terry Lee so thank you in advance.
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
I'm assuming you're having difficulty with the first part of his solution. What he does is use double angles to get rid of the 1 at the front of the numerator and the denominator. I'll do the angles in the numerator as an example and then you should be able to do the rest:

<a href="http://www.codecogs.com/eqnedit.php?latex=cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac{\pi}{14})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac{\pi}{14})" title="cos2\theta =1-2sin^2\theta \\ \therefore cos(\frac{\pi}{7})=1- 2sin^2(\frac{\pi}{14}) \\ sin2\theta =2cos\theta sin\theta \\ \therefore isin(\frac{\pi}{7})=2icos(\frac{\pi}{14})sin(\frac{\pi}{14})" /></a>

When you sub those values in, the 1 will neatly disappear and the same thing will happen with the numerator. Good luck :)
 

Aysce

Well-Known Member
Joined
Jun 24, 2011
Messages
2,394
Gender
Male
HSC
2012
I'm assuming you're having difficulty with the first part of his solution. What he does is use double angles to get rid of the 1 at the front of the numerator and the denominator. I'll do the angles in the numerator as an example and then you should be able to do the rest:



When you sub those values in, the 1 will neatly disappear and the same thing will happen with the numerator. Good luck :)
Thank you <3 This was exactly what I needed.
 

Skuxxgolfer

Member
Joined
Aug 13, 2018
Messages
35
Gender
Male
HSC
N/A
I copy pasted the question and put it into google with the smallest hope that i will find the answer, and by some small miracle we were both looking for the solution to the exact same question, so thank you
 

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

Top