hartnn
  • hartnn
Aaah! new group for trigonometry :) Let me start this by asking to expand sin 5x in powers of sin x. Only for those who have started learning trigonometry and never tried this. Those who already know how to, can give other methods, not discussed here, later.
Trigonometry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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lgbasallote
  • lgbasallote
using the new groups is fun, isn't it?
hartnn
  • hartnn
its just the beginning...... Another similar exercise would be to write sin^5 x in terms of sin5x,sin4x,...
lgbasallote
  • lgbasallote
...i never really learned these stuffs...all i know are the basics of trigonometry....

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lgbasallote
  • lgbasallote
but i do have an idea how to do it......but it's brute force.....is there a shorter way?
hartnn
  • hartnn
there are few ways i am aware of...idk what do u mean by brute force here....u can try to expand here,if u want.
lgbasallote
  • lgbasallote
my idea is using sin (2x) over and over
anonymous
  • anonymous
hmm...the first way that pops into my mind is using:\[e^{ix}=\cos(x)+i\sin(x)\]and being a little clever.
anonymous
  • anonymous
Noting that sin(5x)= Im(cos(5x)+isin(5x))=Im(e^(5ix))=Im((e^ix)^5) =Im((cos(x)+isin(x))^5). So basically, if you see what the imaginary part of:\[(\cos(x)+i \sin(x))^5\]is, you will have a formula for sin(5x).
hartnn
  • hartnn
nice! lets see how new learners try it... maybe they don't even know that e^{ix} identity
ParthKohli
  • ParthKohli
I do know that formula made by Euler, but I didn't know we could apply that here. :P
hartnn
  • hartnn
u can try it parth.... maybe @mathslover also wanna try this....
anonymous
  • anonymous
Here is an example of using Euler's formula to show that:\[\cos(2x)=\cos^2(x)-\sin^2(x)\]\[\sin(2x)=2\sin(x)\cos(x)\]
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mathslover
  • mathslover
Yeah , it is great to see the new pattern : @hartnn can you explain how to do it?

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