What's the fastest way to simplify
^4+\left ( 1+\cos{\frac{\pi}{8}}-i\sin{\frac{\pi}{8}}\right )^4)
without expanding everything/using exact ratios?
Thanks!
without expanding everything/using exact ratios?
Thanks!
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Complex Numbers (1 Viewer)
- Thread starter twinklegal19
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deswa1
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Use double anges to eliminate the 1's. For example, cos(pi/8) becomes 2cos(pi/16)^2 -1 which will 'delete' the 1. Then use double angles on the sin(pi/8)=2sin(pi/16)cos(pi/16). Factorise and then De Moivre's- BOOOOOM.
Nice question though
Nice question though
theind1996
Active Member
Thank you 
Never thought of using double angles haha
Never thought of using double angles haha
deswa1
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Yeah I didn't know how to do that until there was a question in Terry Lee and that's where I learnt that wherever you have 1's in a trig equation, you can just 'delete' them like that. Its a nice trick
theind1996
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iSplicer
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Just an addition: I was taught that, WHENEVER you see (1 + cos[something]) it should set off huge alarm bells. It's VERY likely that you need to use double angles to proceed =). Also, really helps to revise sum to products, and products to sum on the side (for those trickier 4u questions).
SpiralFlex
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This method is long, but what I have done is used the identity repeatedly of
^2-2ab=a^2+b^2)
This may not help with this question but it will certainly help you in the future. Good to find alternate ways than to use the traditional textbook methods.
, \; \overline{z}=(1+\cos \frac{\pi}{8}-i\sin\frac{\pi}{8}))
The expression becomes,
^4=(z^2+(\overline{z})^2)^2-2(z\overline{z})^2)
^2-2z\overline{z})^2-2(z\overline{z})^2)
)^2-2|z|^2)^2-2|z|^4)
This may not help with this question but it will certainly help you in the future. Good to find alternate ways than to use the traditional textbook methods.
The expression becomes,
Nooblet94
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You'll get it from school when we sign out.
theind1996
Active Member
Ah ok, that's good. Hurry the eff up and sign out
Thanks! Just realised how it can also eliminate -1 with double angles as well. Will be sure to remember that next time haha
ThanksThis method is long, but what I have done is used the identity repeatedly of
This may not help with this question but it will certainly help you in the future. Good to find alternate ways than to use the traditional textbook methods.
The expression becomes,
I'm glad to learn any alternative methods, will definitely try to remember this for future problems.