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hartnn
 3 years ago
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.
hartnn
 3 years ago
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.

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lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0using the new groups is fun, isn't it?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.4its just the beginning...... Another similar exercise would be to write sin^5 x in terms of sin5x,sin4x,...

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0...i never really learned these stuffs...all i know are the basics of trigonometry....

lgbasallote
 3 years ago
Best ResponseYou've already chosen the best response.0but i do have an idea how to do it......but it's brute force.....is there a shorter way?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.4there 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
 3 years ago
Best ResponseYou've already chosen the best response.0my idea is using sin (2x) over and over

joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.3hmm...the first way that pops into my mind is using:\[e^{ix}=\cos(x)+i\sin(x)\]and being a little clever.

joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.3Noting 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
 3 years ago
Best ResponseYou've already chosen the best response.4nice! lets see how new learners try it... maybe they don't even know that e^{ix} identity

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I do know that formula made by Euler, but I didn't know we could apply that here. :P

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.4u can try it parth.... maybe @mathslover also wanna try this....

joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.3Here is an example of using Euler's formula to show that:\[\cos(2x)=\cos^2(x)\sin^2(x)\]\[\sin(2x)=2\sin(x)\cos(x)\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah , it is great to see the new pattern : @hartnn can you explain how to do it?
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