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hartnn

  • 2 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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  1. lgbasallote
    • 2 years ago
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    using the new groups is fun, isn't it?

  2. hartnn
    • 2 years ago
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    its just the beginning...... Another similar exercise would be to write sin^5 x in terms of sin5x,sin4x,...

  3. lgbasallote
    • 2 years ago
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    ...i never really learned these stuffs...all i know are the basics of trigonometry....

  4. lgbasallote
    • 2 years ago
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    but i do have an idea how to do it......but it's brute force.....is there a shorter way?

  5. hartnn
    • 2 years ago
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    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.

  6. lgbasallote
    • 2 years ago
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    my idea is using sin (2x) over and over

  7. joemath314159
    • 2 years ago
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    hmm...the first way that pops into my mind is using:\[e^{ix}=\cos(x)+i\sin(x)\]and being a little clever.

  8. joemath314159
    • 2 years ago
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    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).

  9. hartnn
    • 2 years ago
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    nice! lets see how new learners try it... maybe they don't even know that e^{ix} identity

  10. ParthKohli
    • 2 years ago
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    I do know that formula made by Euler, but I didn't know we could apply that here. :P

  11. hartnn
    • 2 years ago
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    u can try it parth.... maybe @mathslover also wanna try this....

  12. joemath314159
    • 2 years ago
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    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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  13. mathslover
    • 2 years ago
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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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