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Complex Numbers (1 Viewer)

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
 

theind1996

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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
Haha fuark,

Strong ability/10.

Did not notice that.
 

deswa1

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Haha fuark,

Strong ability/10.

Did not notice that.
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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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
Does Terry have any other nice tricks in his book?

Perhaps I should buy it...
 

jyu

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Find where 1+z and 1+z'(conjugate) are on the Argand diagram.
 

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



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,





 
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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).
Thanks! Just realised how it can also eliminate -1 with double angles as well. Will be sure to remember that next time haha

This 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,





Thanks :) I'm glad to learn any alternative methods, will definitely try to remember this for future problems.
 

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