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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
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using the new groups is fun, isn't it?
its just the beginning...... Another similar exercise would be to write sin^5 x in terms of sin5x,sin4x,...
...i never really learned these stuffs...all i know are the basics of trigonometry....

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but i do have an idea how to do it......but it's brute force.....is there a shorter way?
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.
my idea is using sin (2x) over and over
hmm...the first way that pops into my mind is using:\[e^{ix}=\cos(x)+i\sin(x)\]and being a little clever.
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).
nice! lets see how new learners try it... maybe they don't even know that e^{ix} identity
I do know that formula made by Euler, but I didn't know we could apply that here. :P
u can try it parth.... maybe @mathslover also wanna try this....
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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Yeah , it is great to see the new pattern : @hartnn can you explain how to do it?

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